Samacheer Kalvi Class 11 Maths Solutions Chapter 9 Limits and Continuity Exercise 9.3

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 9 Limits and Continuity Ex 9.3

Question 1. (a) Find the left and right limits of \( f(x) = \frac{x^2 - 4}{(x^2 + 4x + 4) (x + 3)} \) at \( x = -2 \).
Answer: First, we simplify the function \( f(x) \):
\( f(x) = \frac{x^2 - 4}{(x^2 + 4x + 4)(x + 3)} \)
We factorize the numerator and denominator:
\( f(x) = \frac{(x + 2)(x - 2)}{(x + 2)^2 (x + 3)} \)
\( f(x) = \frac{x - 2}{(x + 2)(x + 3)} \), for \( x \neq -2 \)
Now, we find the left limit of \( f(x) \) at \( x = -2 \):
Let \( x = -2 - h \), where \( h > 0 \). As \( x \to -2 \), \( h \to 0 \).
\( \lim_{x \to -2^-} f(x) = \lim_{h \to 0} \frac{(-2 - h) - 2}{(-2 - h + 2)(-2 - h + 3)} \)
\( = \lim_{h \to 0} \frac{-4 - h}{(-h)(1 - h)} \)
\( = \lim_{h \to 0} \frac{1}{h} \left( \frac{4 + h}{1 - h} \right) \)
Substitute \( h = 0 \):
\( = \frac{1}{0} \left( \frac{4 + 0}{1 - 0} \right) \)
\( = \frac{1}{0} (4) = \infty \)
So, \( \lim_{x \to -2^-} f(x) = \infty \)
Next, we find the right limit of \( f(x) \) at \( x = -2 \):
Let \( x = -2 + h \), where \( h > 0 \). As \( x \to -2 \), \( h \to 0 \).
\( \lim_{x \to -2^+} f(x) = \lim_{h \to 0} \frac{(-2 + h) - 2}{(-2 + h + 2)(-2 + h + 3)} \)
\( = \lim_{h \to 0} \frac{-4 + h}{(h)(1 + h)} \)
\( = \lim_{h \to 0} \frac{1}{h} \left( \frac{-4 + h}{1 + h} \right) \)
Substitute \( h = 0 \):
\( = \frac{1}{0} \left( \frac{-4 + 0}{1 + 0} \right) \)
\( = \frac{1}{0} (-4) = -\infty \)
So, \( \lim_{x \to -2^+} f(x) = -\infty \)
Since the left and right limits are not equal (one is \( \infty \) and the other is \( -\infty \)), the limit of \( f(x) \) at \( x = -2 \) does not exist. This function has a vertical asymptote at \( x = -2 \).
In simple words: We simplify the function first. Then, we check the function's value as \( x \) gets very close to \( -2 \) from both the left and the right sides. The left side goes towards a very large positive number (infinity), and the right side goes towards a very large negative number (negative infinity). Because they are different, the overall limit at \( -2 \) does not exist.

🎯 Exam Tip: Always simplify the rational function by factoring before evaluating limits, especially when direct substitution leads to an indeterminate form like \( \frac{0}{0} \). This helps reveal vertical asymptotes or holes.

Question 1. (b) Find the left and right limits of \( f(x) = \tan x \) at \( x = \frac{\pi}{2} \).
Answer: We find the left limit of \( f(x) \) at \( x = \frac{\pi}{2} \):
Let \( x = \frac{\pi}{2} - h \), where \( h > 0 \). As \( x \to \frac{\pi}{2}^- \), \( h \to 0 \).
\( \lim_{x \to \frac{\pi}{2}^-} (\tan x) = \lim_{h \to 0} \tan \left( \frac{\pi}{2} - h \right) \)
Since \( \tan(\frac{\pi}{2} - h) = \cot h \):
\( = \lim_{h \to 0} \cot h \)
As \( h \to 0 \), \( \cot h \to \infty \).
So, \( \lim_{x \to \frac{\pi}{2}^-} (\tan x) = \infty \)
Next, we find the right limit of \( f(x) \) at \( x = \frac{\pi}{2} \):
Let \( x = \frac{\pi}{2} + h \), where \( h > 0 \). As \( x \to \frac{\pi}{2}^+ \), \( h \to 0 \).
\( \lim_{x \to \frac{\pi}{2}^+} (\tan x) = \lim_{h \to 0} \tan \left( \frac{\pi}{2} + h \right) \)
Since \( \tan(\frac{\pi}{2} + h) = -\cot h \):
\( = \lim_{h \to 0} (-\cot h) \)
As \( h \to 0 \), \( -\cot h \to -\infty \).
So, \( \lim_{x \to \frac{\pi}{2}^+} (\tan x) = -\infty \)
Since the left limit is \( \infty \) and the right limit is \( -\infty \), the limit of \( f(x) = \tan x \) at \( x = \frac{\pi}{2} \) does not exist. The tangent function has vertical asymptotes at \( x = \frac{\pi}{2} + n\pi \).
In simple words: We look at the tangent function as \( x \) gets very close to \( \frac{\pi}{2} \) from both sides. From the left, the value shoots up to positive infinity. From the right, it drops down to negative infinity. Because these are different, the limit does not exist at \( \frac{\pi}{2} \).

🎯 Exam Tip: Remember the graphs and properties of trigonometric functions, especially where they have asymptotes, as this directly affects the existence of limits. Tangent has infinite discontinuities.

Question 2. Evaluate the following limits:
\( \lim_{x \to 3} \frac{x^2 - 9}{x^2(x^2 - 6x + 9)} \)
Answer: We need to evaluate the limit \( \lim_{x \to 3} \frac{x^2 - 9}{x^2(x^2 - 6x + 9)} \).
First, we simplify the expression by factoring the numerator and denominator:
Numerator: \( x^2 - 9 = (x - 3)(x + 3) \)
Denominator: \( x^2 - 6x + 9 = (x - 3)^2 \)
So, the function becomes:
\( f(x) = \frac{(x - 3)(x + 3)}{x^2 (x - 3)^2} = \frac{x + 3}{x^2 (x - 3)} \), for \( x \neq 3 \)
Now, we find the left limit of \( f(x) \) at \( x = 3 \):
Let \( x = 3 - h \), where \( h > 0 \). As \( x \to 3^- \), \( h \to 0 \).
\( \lim_{x \to 3^-} f(x) = \lim_{h \to 0} \frac{(3 - h) + 3}{(3 - h)^2 ((3 - h) - 3)} \)
\( = \lim_{h \to 0} \frac{6 - h}{(3 - h)^2 (-h)} \)
Substitute \( h = 0 \):
\( = \frac{6 - 0}{(3 - 0)^2 (-0)} = \frac{6}{9 \cdot 0^-} = -\infty \)
Next, we find the right limit of \( f(x) \) at \( x = 3 \):
Let \( x = 3 + h \), where \( h > 0 \). As \( x \to 3^+ \), \( h \to 0 \).
\( \lim_{x \to 3^+} f(x) = \lim_{h \to 0} \frac{(3 + h) + 3}{(3 + h)^2 ((3 + h) - 3)} \)
\( = \lim_{h \to 0} \frac{6 + h}{(3 + h)^2 (h)} \)
Substitute \( h = 0 \):
\( = \frac{6 + 0}{(3 + 0)^2 (0)} = \frac{6}{9 \cdot 0^+} = \infty \)
Since the left limit is \( -\infty \) and the right limit is \( \infty \), the limit of \( f(x) \) at \( x = 3 \) does not exist. This indicates a vertical asymptote at \( x = 3 \).
In simple words: First, we factor the top and bottom parts of the fraction and cancel out common terms. Then, we check the value of the function as \( x \) gets very close to 3 from the left side and the right side. From the left, it goes to negative infinity, and from the right, it goes to positive infinity. Since these are different, the limit at \( x = 3 \) does not exist.

🎯 Exam Tip: When evaluating limits where direct substitution leads to \( \frac{k}{0} \) (where \( k \neq 0 \)), always check left and right limits separately to determine if the limit is \( \infty \), \( -\infty \), or does not exist. This is crucial for understanding behavior near vertical asymptotes.

Question 3. Evaluate the following limits:
\( \lim_{x \to \infty} \left( \frac{3}{x - 2} - \frac{2x + 11}{x^2 + x - 6} \right) \)
Answer: We need to evaluate the limit \( \lim_{x \to \infty} \left( \frac{3}{x - 2} - \frac{2x + 11}{x^2 + x - 6} \right) \).
First, we find a common denominator for the two fractions. The denominator of the second term can be factored: \( x^2 + x - 6 = (x + 3)(x - 2) \).
So, the expression becomes:
\( \lim_{x \to \infty} \left( \frac{3}{x - 2} - \frac{2x + 11}{(x + 3)(x - 2)} \right) \)
Combine the fractions:
\( = \lim_{x \to \infty} \frac{3(x + 3) - (2x + 11)}{(x + 3)(x - 2)} \)
\( = \lim_{x \to \infty} \frac{3x + 9 - 2x - 11}{(x + 3)(x - 2)} \)
\( = \lim_{x \to \infty} \frac{x - 2}{(x + 3)(x - 2)} \)
Cancel out the common factor \( (x - 2) \), assuming \( x \neq 2 \):
\( = \lim_{x \to \infty} \frac{1}{x + 3} \)
Now, substitute \( x = \infty \):
\( = \frac{1}{\infty + 3} = \frac{1}{\infty} = 0 \)
Therefore, the limit is 0.
In simple words: We first combine the two fractions into one by finding a common bottom part. Then we simplify the new fraction by canceling out anything that appears on both the top and bottom. Finally, as \( x \) gets very large (goes to infinity), the fraction becomes 1 divided by a very large number, which means it gets closer and closer to zero.

🎯 Exam Tip: When dealing with limits at infinity for rational functions or expressions involving fractions, always combine terms and simplify first. Then, divide both the numerator and denominator by the highest power of \( x \) present to easily evaluate the limit.

Question 4. Evaluate the following limits:
\( \lim_{x \to \infty} \frac{x^3 + x}{x^4 - 3x^2 + 1} \)
Answer: We need to evaluate the limit \( \lim_{x \to \infty} \frac{x^3 + x}{x^4 - 3x^2 + 1} \).
To find the limit as \( x \to \infty \), we divide both the numerator and the denominator by the highest power of \( x \) in the denominator, which is \( x^4 \).
\( = \lim_{x \to \infty} \frac{\frac{x^3}{x^4} + \frac{x}{x^4}}{\frac{x^4}{x^4} - \frac{3x^2}{x^4} + \frac{1}{x^4}} \)
\( = \lim_{x \to \infty} \frac{\frac{1}{x} + \frac{1}{x^3}}{1 - \frac{3}{x^2} + \frac{1}{x^4}} \)
As \( x \to \infty \), terms like \( \frac{1}{x}, \frac{1}{x^3}, \frac{3}{x^2}, \frac{1}{x^4} \) all approach 0.
So, we substitute these values:
\( = \frac{0 + 0}{1 - 0 + 0} \)
\( = \frac{0}{1} = 0 \)
Therefore, the limit is 0.
In simple words: To find what happens to the fraction as \( x \) becomes extremely large, we divide every part of the top and bottom by the biggest power of \( x \) found at the bottom. As \( x \) gets huge, any term like "1 divided by \( x \)" becomes zero. After doing this, we see that the top part becomes zero and the bottom part becomes one. So, the whole fraction gets closer to zero.

🎯 Exam Tip: For limits at infinity of rational functions, compare the highest powers of \( x \) in the numerator and denominator. If the degree of the denominator is greater, the limit is 0. If degrees are equal, it's the ratio of leading coefficients. If the numerator degree is greater, the limit is \( \infty \) or \( -\infty \).

Question 5. Evaluate the following limits:
\( \lim_{x \to \infty} \frac{x^4 - 5x}{x^2 - 3x + 1} \)
Answer: We need to evaluate the limit \( \lim_{x \to \infty} \frac{x^4 - 5x}{x^2 - 3x + 1} \).
To find the limit as \( x \to \infty \), we divide both the numerator and the denominator by the highest power of \( x \) in the denominator, which is \( x^2 \).
\( = \lim_{x \to \infty} \frac{\frac{x^4}{x^2} - \frac{5x}{x^2}}{\frac{x^2}{x^2} - \frac{3x}{x^2} + \frac{1}{x^2}} \)
\( = \lim_{x \to \infty} \frac{x^2 - \frac{5}{x}}{1 - \frac{3}{x} + \frac{1}{x^2}} \)
As \( x \to \infty \), terms like \( \frac{5}{x}, \frac{3}{x}, \frac{1}{x^2} \) all approach 0. However, \( x^2 \) approaches \( \infty \).
So, we substitute these values:
\( = \frac{\infty - 0}{1 - 0 + 0} \)
\( = \frac{\infty}{1} = \infty \)
Therefore, the limit is \( \infty \).
In simple words: When \( x \) becomes very large, we divide all parts of the fraction by the biggest power of \( x \) in the bottom part. This shows us that the top part grows infinitely large (like \( x^2 \)), while the bottom part approaches 1. So, the whole fraction becomes infinitely large.

🎯 Exam Tip: If the degree of the numerator is greater than the degree of the denominator in a rational function as \( x \to \infty \), the limit will be either \( \infty \) or \( -\infty \). The sign depends on the leading coefficients of the numerator and denominator.

Question 6. Evaluate the following limits:
\( \lim_{x \to \infty} \frac{1 + x - 3x^3}{1 + x^2 + 3x^3} \)
Answer: We need to evaluate the limit \( \lim_{x \to \infty} \frac{1 + x - 3x^3}{1 + x^2 + 3x^3} \).
To find the limit as \( x \to \infty \), we divide both the numerator and the denominator by the highest power of \( x \) in the denominator, which is \( x^3 \).
\( = \lim_{x \to \infty} \frac{\frac{1}{x^3} + \frac{x}{x^3} - \frac{3x^3}{x^3}}{\frac{1}{x^3} + \frac{x^2}{x^3} + \frac{3x^3}{x^3}} \)
\( = \lim_{x \to \infty} \frac{\frac{1}{x^3} + \frac{1}{x^2} - 3}{\frac{1}{x^3} + \frac{1}{x} + 3} \)
As \( x \to \infty \), terms like \( \frac{1}{x^3}, \frac{1}{x^2}, \frac{1}{x} \) all approach 0.
So, we substitute these values:
\( = \frac{0 + 0 - 3}{0 + 0 + 3} \)
\( = \frac{-3}{3} = -1 \)
Therefore, the limit is -1.
In simple words: For a fraction where \( x \) is getting very big, we divide every part of the top and bottom by the largest power of \( x \) in the bottom. All parts that look like "a number divided by \( x \)" will become zero. We are left with just the numbers that were multiplied by the highest power of \( x \). The result is the ratio of these numbers, which is -1.

🎯 Exam Tip: When the degrees of the numerator and denominator are the same for limits at infinity, the limit is simply the ratio of their leading coefficients. This is a quick shortcut to check your work after a full calculation.

Question 7. Evaluate the following limits:
\( \lim_{x \to \infty} \left( \frac{x^3}{2x^2 - 1} - \frac{x^2}{2x + 1} \right) \)
Answer: We need to evaluate the limit \( \lim_{x \to \infty} \left( \frac{x^3}{2x^2 - 1} - \frac{x^2}{2x + 1} \right) \).
First, we find a common denominator for the two fractions:
\( = \lim_{x \to \infty} \frac{x^3(2x + 1) - x^2(2x^2 - 1)}{(2x^2 - 1)(2x + 1)} \)
Expand the numerator and denominator:
\( = \lim_{x \to \infty} \frac{2x^4 + x^3 - (2x^4 - x^2)}{4x^3 + 2x^2 - 2x - 1} \)
\( = \lim_{x \to \infty} \frac{2x^4 + x^3 - 2x^4 + x^2}{4x^3 + 2x^2 - 2x - 1} \)
\( = \lim_{x \to \infty} \frac{x^3 + x^2}{4x^3 + 2x^2 - 2x - 1} \)
Now, we divide both the numerator and the denominator by the highest power of \( x \) in the denominator, which is \( x^3 \).
\( = \lim_{x \to \infty} \frac{\frac{x^3}{x^3} + \frac{x^2}{x^3}}{\frac{4x^3}{x^3} + \frac{2x^2}{x^3} - \frac{2x}{x^3} - \frac{1}{x^3}} \)
\( = \lim_{x \to \infty} \frac{1 + \frac{1}{x}}{4 + \frac{2}{x} - \frac{2}{x^2} - \frac{1}{x^3}} \)
As \( x \to \infty \), terms like \( \frac{1}{x}, \frac{2}{x}, \frac{2}{x^2}, \frac{1}{x^3} \) all approach 0.
So, we substitute these values:
\( = \frac{1 + 0}{4 + 0 - 0 - 0} \)
\( = \frac{1}{4} \)
Therefore, the limit is \( \frac{1}{4} \).
In simple words: First, we combine the two fractions by finding a common bottom part. This creates one big fraction. Then, we simplify the top and bottom of this big fraction. Since \( x \) is going to infinity, we divide every term by the highest power of \( x \) we see. All parts that become a number divided by \( x \) (or \( x \) squared, etc.) will turn into zero. What's left gives us the final answer, which is \( \frac{1}{4} \).

🎯 Exam Tip: For expressions involving differences of rational functions as \( x \to \infty \), always combine them into a single fraction first. Then, apply the standard technique of dividing by the highest power of \( x \) in the denominator to find the limit.

Question 8. Show that
(i) \( \lim_{n \to \infty} \frac{1 + 2 + 3 + .... + n}{3n^2 + 7n + 2} = \frac{1}{6} \)
Answer: We need to show that \( \lim_{n \to \infty} \frac{1 + 2 + 3 + .... + n}{3n^2 + 7n + 2} = \frac{1}{6} \).
First, we use the formula for the sum of the first \( n \) natural numbers: \( 1 + 2 + 3 + .... + n = \frac{n(n + 1)}{2} \).
Substitute this into the expression:
\( = \lim_{n \to \infty} \frac{\frac{n(n + 1)}{2}}{3n^2 + 7n + 2} \)
\( = \lim_{n \to \infty} \frac{n(n + 1)}{2(3n^2 + 7n + 2)} \)
\( = \lim_{n \to \infty} \frac{n^2 + n}{6n^2 + 14n + 4} \)
Now, we divide both the numerator and the denominator by the highest power of \( n \) in the denominator, which is \( n^2 \).
\( = \lim_{n \to \infty} \frac{\frac{n^2}{n^2} + \frac{n}{n^2}}{\frac{6n^2}{n^2} + \frac{14n}{n^2} + \frac{4}{n^2}} \)
\( = \lim_{n \to \infty} \frac{1 + \frac{1}{n}}{6 + \frac{14}{n} + \frac{4}{n^2}} \)
As \( n \to \infty \), terms like \( \frac{1}{n}, \frac{14}{n}, \frac{4}{n^2} \) all approach 0.
So, we substitute these values:
\( = \frac{1 + 0}{6 + 0 + 0} \)
\( = \frac{1}{6} \)
Thus, it is shown that \( \lim_{n \to \infty} \frac{1 + 2 + 3 + .... + n}{3n^2 + 7n + 2} = \frac{1}{6} \).
In simple words: We replace the sum of numbers from 1 to \( n \) with its formula, \( \frac{n(n+1)}{2} \). Then, we simplify the fraction. As \( n \) becomes very large, we divide every part by the highest power of \( n \). This makes many terms turn into zero, leaving us with just the main numbers on top and bottom, which gives \( \frac{1}{6} \).

🎯 Exam Tip: Always recall and correctly apply the sum formulas for arithmetic progressions (like \( \sum n \)) and sums of squares (like \( \sum n^2 \)) when they appear in limit problems. Simplifying the expression before taking the limit is key.

Question 8. (ii) Show that
\( \lim_{n \to \infty} \frac{1^2 + 2^2 + .... + (3n)^2}{(1 + 2 + ..... + 5n) (2n + 3)} = \frac{9}{25} \)
Answer: We need to show that \( \lim_{n \to \infty} \frac{1^2 + 2^2 + .... + (3n)^2}{(1 + 2 + ..... + 5n) (2n + 3)} = \frac{9}{25} \).
First, we use the sum formulas:
Sum of squares: \( 1^2 + 2^2 + .... + k^2 = \frac{k(k + 1)(2k + 1)}{6} \)
Here, \( k = 3n \), so the numerator is \( \frac{3n(3n + 1)(2(3n) + 1)}{6} = \frac{3n(3n + 1)(6n + 1)}{6} \).
Sum of natural numbers: \( 1 + 2 + .... + k = \frac{k(k + 1)}{2} \)
Here, \( k = 5n \), so the sum in the denominator is \( \frac{5n(5n + 1)}{2} \).
Substitute these into the expression:
\( = \lim_{n \to \infty} \frac{\frac{3n(3n + 1)(6n + 1)}{6}}{\frac{5n(5n + 1)}{2} (2n + 3)} \)
Rearrange the terms:
\( = \lim_{n \to \infty} \frac{3n(3n + 1)(6n + 1)}{6} \cdot \frac{2}{5n(5n + 1)(2n + 3)} \)
\( = \lim_{n \to \infty} \frac{3n(3n + 1)(6n + 1)}{3 \cdot 5n(5n + 1)(2n + 3)} \)
Cancel \( 3n \) from numerator and denominator:
\( = \lim_{n \to \infty} \frac{(3n + 1)(6n + 1)}{5(5n + 1)(2n + 3)} \)
Now, divide both the numerator and the denominator by \( n^2 \) (the highest power of \( n \) in the denominator if you expand it).
\( = \lim_{n \to \infty} \frac{\frac{3n+1}{n} \cdot \frac{6n+1}{n}}{5 \cdot \frac{5n+1}{n} \cdot \frac{2n+3}{n}} \)
\( = \lim_{n \to \infty} \frac{(3 + \frac{1}{n})(6 + \frac{1}{n})}{5(5 + \frac{1}{n})(2 + \frac{3}{n})} \)
As \( n \to \infty \), terms like \( \frac{1}{n}, \frac{3}{n} \) all approach 0.
So, we substitute these values:
\( = \frac{(3 + 0)(6 + 0)}{5(5 + 0)(2 + 0)} \)
\( = \frac{3 \cdot 6}{5 \cdot 5 \cdot 2} = \frac{18}{50} = \frac{9}{25} \)
Thus, it is shown that \( \lim_{n \to \infty} \frac{1^2 + 2^2 + .... + (3n)^2}{(1 + 2 + ..... + 5n) (2n + 3)} = \frac{9}{25} \).
In simple words: We replace the sum of squares and the sum of natural numbers with their respective formulas. Then, we simplify the complex fraction by canceling terms and rearranging. Finally, as \( n \) gets infinitely large, we divide by the highest power of \( n \) to find the limit. This simplifies to \( \frac{9}{25} \).

🎯 Exam Tip: When simplifying complex fractions involving series, be careful with the multiplication of denominators and canceling common factors. Pay close attention to the highest degree terms to predict the limit quickly.

Question 8. (iii) Show that
\( \lim_{n \to \infty} \left[ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} +....+ \frac{1}{n(n+1)} \right] = 1 \)
Answer: We need to show that \( \lim_{n \to \infty} \left[ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} +....+ \frac{1}{n(n+1)} \right] = 1 \).
This is a sum involving partial fractions. Each term \( \frac{1}{k(k+1)} \) can be written as \( \frac{1}{k} - \frac{1}{k+1} \).
Let \( T_n = \frac{1}{n(n+1)} \). So, \( T_n = \frac{1}{n} - \frac{1}{n+1} \).
Let's write out the first few terms of the sum:
\( T_1 = \frac{1}{1} - \frac{1}{2} \)
\( T_2 = \frac{1}{2} - \frac{1}{3} \)
\( T_3 = \frac{1}{3} - \frac{1}{4} \)
...
\( T_n = \frac{1}{n} - \frac{1}{n+1} \)
Now, we find the sum \( S_n = T_1 + T_2 + .... + T_n \):
\( S_n = \left( 1 - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + .... + \left( \frac{1}{n} - \frac{1}{n+1} \right) \)
This is a telescoping sum, where intermediate terms cancel out:
\( S_n = 1 - \frac{1}{n+1} \)
Now, we evaluate the limit of \( S_n \) as \( n \to \infty \):
\( \lim_{n \to \infty} S_n = \lim_{n \to \infty} \left( 1 - \frac{1}{n+1} \right) \)
As \( n \to \infty \), \( \frac{1}{n+1} \) approaches 0.
\( = 1 - 0 = 1 \)
Thus, it is shown that \( \lim_{n \to \infty} \left[ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} +....+ \frac{1}{n(n+1)} \right] = 1 \).
In simple words: Each part of the sum can be broken down into two simpler fractions, like \( \frac{1}{k} - \frac{1}{k+1} \). When we add all these parts together, most of them cancel each other out, leaving only the first term and the very last term. As \( n \) gets infinitely large, the last term becomes zero, so the whole sum approaches 1.

🎯 Exam Tip: Recognize telescoping series. The ability to express terms as a difference allows for significant cancellation, simplifying the sum to just the first and last terms, making the limit calculation straightforward.

Question 9. An important problem in fishery science is to estimate the number of fish in streams and use this information to predict the number of mature fish or "recruits" that will return to the rivers during the reproductive period. If S is the number of spawners and R the number of recruits, "Beverton-Holt spawner recruit function" is \( R(S) = \frac{S}{\alpha S + \beta} \) where \( \alpha \) and \( \beta \) are positive constants. Show that this function predicts approximately constant recruitment when the number of spawners is sufficiently large.
Answer: Given the Beverton-Holt spawner recruit function: \( R(S) = \frac{S}{\alpha S + \beta} \)
Here, \( S \) is the number of spawners, and \( R \) is the number of recruits. \( \alpha \) and \( \beta \) are positive constants.
We need to show that recruitment is approximately constant when the number of spawners \( S \) is sufficiently large, which means we need to evaluate the limit of \( R(S) \) as \( S \to \infty \).
\( \lim_{S \to \infty} R(S) = \lim_{S \to \infty} \frac{S}{\alpha S + \beta} \)
To evaluate this limit, we divide both the numerator and the denominator by the highest power of \( S \) in the denominator, which is \( S \).
\( = \lim_{S \to \infty} \frac{\frac{S}{S}}{\frac{\alpha S}{S} + \frac{\beta}{S}} \)
\( = \lim_{S \to \infty} \frac{1}{\alpha + \frac{\beta}{S}} \)
As \( S \to \infty \), the term \( \frac{\beta}{S} \) approaches 0.
So, we substitute this value:
\( = \frac{1}{\alpha + 0} \)
\( = \frac{1}{\alpha} \)
Since \( \alpha \) is a positive constant, \( \frac{1}{\alpha} \) is also a constant value. Therefore, when the number of spawners \( S \) is sufficiently large, the number of recruits \( R(S) \) approaches the constant value \( \frac{1}{\alpha} \). This shows that the function predicts approximately constant recruitment in such a scenario.
In simple words: We are looking at what happens to the number of new fish (recruits) when there are a huge number of parent fish (spawners). We use a special function and see its value as the number of parent fish goes to infinity. We divide the top and bottom of the function by 'S' (number of spawners). This shows that the number of new fish will settle down to a fixed number, \( \frac{1}{\alpha} \), meaning the recruitment becomes almost constant, regardless of how many more spawners there are.

🎯 Exam Tip: For real-world application problems involving limits at infinity, the interpretation of the limit is as important as the calculation. Explain what the constant value means in the context of the problem, like saturation or a stable population size.

Question 10. A tank contains 5000 litres of pure water. Brine (very salty water) that contains 30 grams of salt per litre of water is pumped into the tank at a rate of 25 litres per minute. The concentration of saltwater after t minutes (in grams per litre) is \( C(t) = \frac{30 t}{200+t} \). What happens to the concentrations as \( t \to \infty \)?
Answer: Given the concentration of saltwater after \( t \) minutes is \( C(t) = \frac{30t}{200+t} \).
We want to find what happens to the concentration as \( t \to \infty \), which means we need to evaluate the limit of \( C(t) \) as \( t \) approaches infinity.
\( \lim_{t \to \infty} C(t) = \lim_{t \to \infty} \frac{30t}{200+t} \)
To evaluate this limit, we divide both the numerator and the denominator by the highest power of \( t \) in the denominator, which is \( t \).
\( = \lim_{t \to \infty} \frac{\frac{30t}{t}}{\frac{200}{t} + \frac{t}{t}} \)
\( = \lim_{t \to \infty} \frac{30}{\frac{200}{t} + 1} \)
As \( t \to \infty \), the term \( \frac{200}{t} \) approaches 0.
So, we substitute this value:
\( = \frac{30}{0 + 1} \)
\( = \frac{30}{1} = 30 \)
Therefore, as \( t \to \infty \), the concentration of saltwater approaches 30 grams per litre. This means that over a very long time, the concentration of salt in the tank will stabilize at 30 grams per litre, which is the concentration of the incoming brine.
In simple words: We are looking at the salt concentration in the tank over a very long time. As time \( t \) goes to infinity, we divide every part of the concentration formula by \( t \). This makes the part \( \frac{200}{t} \) become zero. So, the concentration gets closer and closer to 30 grams per litre, which is the concentration of the salty water being added.

🎯 Exam Tip: For mixture problems involving limits at infinity, the limit often represents the equilibrium or steady-state concentration. Ensure you interpret the numerical result correctly in the context of the physical situation, explaining what it means for the concentration over time.

Question 10. A tank contains 5000 litres of pure water. Brine (very salty water) that contains 30 grams of salt per litre of water is pumped into the tank at a rate of 25 litres per minute. The concentration of saltwater after t minutes (in grams per litre) is \( C(t) = \frac{30 t}{200+t} \). What happens to the concentrations as \( t \to \infty \)?
Answer: The concentration of saltwater after \( t \) minutes is given by the formula \( C(t) = \frac{30 t}{200+t} \). To find out what happens to the concentration as time \( t \) approaches infinity, we need to calculate the limit of \( C(t) \) as \( t \to \infty \).
\( \lim_{t \to \infty} C(t) = \lim_{t \to \infty} \frac{30 t}{200 + t} \)
To evaluate this limit, we divide both the numerator and the denominator by \( t \), which is the highest power of \( t \) in the denominator:
\( = \lim_{t \to \infty} \frac{\frac{30 t}{t}}{\frac{200}{t} + \frac{t}{t}} \)
\( = \lim_{t \to \infty} \frac{30}{\frac{200}{t} + 1} \)
As \( t \) gets very large (approaches infinity), the term \( \frac{200}{t} \) becomes very small and approaches 0. This shows that the initial volume of pure water becomes less significant over a long period.
\( = \frac{30}{0 + 1} \)
\( = 30 \)
So, as \( t \to \infty \), the concentration of saltwater approaches 30 grams per litre.
In simple words: As a very long time passes, the amount of salt per litre in the tank gets closer and closer to 30 grams. It reaches a stable level because new salty water keeps flowing in.

🎯 Exam Tip: When finding limits at infinity for fractions involving variables, divide both the top and bottom by the highest power of the variable to simplify the expression and evaluate it easily.