OP Malhotra Class 11 Maths Solutions Chapter 18 Limits Exercise 18 (B)

Question 1. \( \text{Lim}_{x \to 0} (7x^2-5x+1) \)
Answer: To find the limit, we directly substitute \( x = 0 \) into the expression.
\( \text{Lt}_{x \to 0} (7x^2 - 5x + 1) = 7 \times (0)^2 - 5 \times 0 + 1 \)
\( = 7 \times 0 - 0 + 1 \)
\( = 0 - 0 + 1 \)
\( = 1 \)
In simple words: To solve this, just replace every 'x' with '0' in the given math problem. After calculating, the final answer is 1.

🎯 Exam Tip: For polynomial functions, the limit as x approaches a finite number can often be found by direct substitution. Always check for indeterminate forms (like 0/0) first, but for simple polynomials, direct substitution is usually enough.

Question 2. \( \text{Lim}_{x \to 1} \frac{x-1}{x+1} \)
Answer: We can find this limit by substituting \( x = 1 \) directly into the expression, as it does not result in an indeterminate form.
\( \text{Lt}_{x \to 1} \frac{x-1}{x+1} = \frac{1-1}{1+1} \)
\( = \frac{0}{2} \)
\( = 0 \)
In simple words: Replace 'x' with '1' in the top and bottom parts of the fraction. The top becomes 0, and the bottom becomes 2, so the final answer is 0.

🎯 Exam Tip: When dealing with limits of rational functions (fractions with polynomials), direct substitution is valid as long as the denominator does not become zero. If it does, further simplification (like factoring) might be needed.

Question 3. \( \text{Lim}_{x \to 2} \frac{x^2+5x+6}{2x^2-3x} \)
Answer: We substitute \( x = 2 \) into the numerator and the denominator directly because the denominator will not be zero.
\( \text{Lt}_{x \to 2} \frac{x^2+5x+6}{2x^2-3x} = \frac{(2)^2+5 \times 2+6}{2 \times (2)^2-3 \times 2} \)
\( = \frac{4+10+6}{2 \times 4-6} \)
\( = \frac{20}{8-6} \)
\( = \frac{20}{2} \)
\( = 10 \)
In simple words: Put '2' in place of 'x' everywhere in the fraction. Calculate the top part and the bottom part separately, then divide them to get the answer, which is 10.

🎯 Exam Tip: Always evaluate the denominator first when directly substituting into a rational function limit. If the denominator is non-zero, direct substitution is the simplest method. If the denominator is zero, you must simplify the expression further.

Question 4. \( \text{Lim}_{x \to 0} \frac{ax+b}{cx+d} \)
Answer: For this rational function, we can substitute \( x = 0 \) directly into the expression, assuming \( d \neq 0 \).
\( \text{Lt}_{x \to 0} \frac{ax+b}{cx+d} = \frac{a \times 0+b}{c \times 0+d} \)
\( = \frac{0+b}{0+d} \)
\( = \frac{b}{d} \)
In simple words: Replace 'x' with '0' in the fraction. The terms with 'x' will disappear, leaving 'b' on top and 'd' on the bottom. So the answer is b/d.

🎯 Exam Tip: Direct substitution is effective for limits of rational functions if the denominator does not become zero. If the problem doesn't state it, assume 'd' is not zero for the limit to exist as a finite number.

Question 5. \( \text{Lim}_{x \to 5^+} \frac{x-5}{|x-5|} \)
Answer: We need to evaluate the right-hand limit as \( x \) approaches 5. When \( x \to 5^+ \), it means \( x \) is slightly greater than 5.
\( \implies \) Therefore, \( x-5 \) will be a small positive number.
\( \implies \) For a positive number, the absolute value \( |x-5| \) is simply \( (x-5) \).
\( \text{Lt}_{x \to 5^+} \frac{x-5}{|x-5|} = \text{Lt}_{x \to 5^+} \frac{x-5}{x-5} \)
\( = \text{Lt}_{x \to 5^+} 1 \)
\( = 1 \)
In simple words: When 'x' is just a tiny bit bigger than 5, then 'x minus 5' will be a small positive number. The absolute value of a positive number is just the number itself. So, the top and bottom of the fraction become the same, meaning the answer is 1.

🎯 Exam Tip: For limits involving absolute values, first determine if the expression inside the absolute value is positive or negative based on the direction of the limit (e.g., \( x \to a^+ \) or \( x \to a^- \)). This helps in removing the absolute value sign correctly.

Question 6. \( \text{Lim}_{x \to -5^-} \frac{x+5}{|x+5|} \)
Answer: We are evaluating the left-hand limit as \( x \) approaches \( -5 \).
\( \implies \) When \( x \to -5^- \), it means \( x \) is a number slightly less than \( -5 \) (e.g., \( -5.0001 \)).
\( \implies \) This makes the expression \( x+5 \) a small negative number.
\( \implies \) For a negative number, the absolute value \( |x+5| \) is \( -(x+5) \).
\( \text{Lt}_{x \to -5^-} \frac{x+5}{|x+5|} = \text{Lt}_{x \to -5^-} \frac{x+5}{-(x+5)} \)
\( = \text{Lt}_{x \to -5^-} (-1) \)
\( = -1 \)
In simple words: If 'x' is a little smaller than -5, then 'x plus 5' will be a small negative number. The absolute value of a negative number means we change its sign to positive. So, the fraction becomes a number divided by its negative, which always equals -1.

🎯 Exam Tip: Carefully determine the sign of the expression inside the absolute value when approaching a point from the left or right. If \( f(x) < 0 \), then \( |f(x)| = -f(x) \), and if \( f(x) > 0 \), then \( |f(x)| = f(x) \).

Question 7. \( \text{Lim}_{x \to 5^+} (x-[x]) \)
Answer: We need to find the right-hand limit of the expression \( x-[x] \) as \( x \) approaches 5.
\( \implies \) Let \( x = 5+h \), where \( h \) is a very small positive number and \( h \to 0^+ \).
\( \implies \) Then, the greatest integer function \( [x] \) becomes \( [5+h] \).
\( \implies \) Since \( h \) is a small positive number, \( [5+h] = 5 \). This is because the greatest integer less than or equal to a number like 5.0001 is 5.
\( \text{Lt}_{x \to 5^+} (x-[x]) = \text{Lt}_{h \to 0^+} ((5+h)-[5+h]) \)
\( = \text{Lt}_{h \to 0^+} (5+h-5) \)
\( = \text{Lt}_{h \to 0^+} h \)
\( = 0 \)
In simple words: When 'x' is just slightly more than 5, like 5.001, then the greatest integer less than or equal to 'x' is 5. So, the problem becomes (5 + a tiny bit) minus 5, which leaves just that tiny bit. As the tiny bit becomes zero, the answer is 0.

🎯 Exam Tip: Remember that for the greatest integer function \( [x] \), if \( n \) is an integer, then \( \text{Lim}_{x \to n^+} [x] = n \) and \( \text{Lim}_{x \to n^-} [x] = n-1 \). This understanding is critical for evaluating limits involving this function.

Question 8. \( \text{Lim}_{x \to 1^-} \frac{x^2-1}{|x-1|} \)
Answer: We need to find the left-hand limit as \( x \) approaches 1.
\( \implies \) Let \( x = 1-h \), where \( h \) is a very small positive number and \( h \to 0^+ \).
\( \implies \) Then \( x-1 = (1-h)-1 = -h \).
\( \implies \) Since \( h > 0 \), \( -h \) is a negative number. So, \( |x-1| = |-h| = -(-h) = h \).
\( \text{Lt}_{x \to 1^-} \frac{x^2-1}{|x-1|} = \text{Lt}_{h \to 0^+} \frac{(1-h)^2-1}{h} \)
\( = \text{Lt}_{h \to 0^+} \frac{1-2h+h^2-1}{h} \)
\( = \text{Lt}_{h \to 0^+} \frac{h^2-2h}{h} \)
\( = \text{Lt}_{h \to 0^+} \frac{h(h-2)}{h} \)
\( = \text{Lt}_{h \to 0^+} (h-2) \)
\( = 0-2 \)
\( = -2 \)
In simple words: When 'x' is slightly less than 1, like 0.999, then 'x minus 1' is a small negative number. The absolute value of this negative number makes it positive. We can factorize the top part and cancel out a common term with the bottom. After that, put 0 for 'h' to get -2.

🎯 Exam Tip: When \( x \to a^- \), let \( x = a-h \) where \( h \to 0^+ \). When \( x \to a^+ \), let \( x = a+h \) where \( h \to 0^+ \). This substitution helps correctly handle absolute values and greatest integer functions by turning a two-sided limit into a right-sided limit at 0.

Question 9. Show that \( \text{Lim}_{x \to 2} \log_{10} \{x^6 + \sqrt{x^2 + 1292}\} = 2 \).
Answer: To show the limit, we substitute \( x = 2 \) directly into the expression inside the logarithm, since the function is continuous for positive arguments.
\( \text{Lt}_{x \to 2} \log_{10} \{x^6 + \sqrt{x^2 + 1292}\} \)
\( = \log_{10} \{ (2)^6 + \sqrt{(2)^2 + 1292} \} \)
\( = \log_{10} \{ 64 + \sqrt{4 + 1292} \} \)
\( = \log_{10} \{ 64 + \sqrt{1296} \} \)
\( = \log_{10} \{ 64 + 36 \} \)
\( = \log_{10} 100 \)
\( = \log_{10} 10^2 \)
Using the logarithm property \( \log_a a^b = b \), we get:
\( = 2 \times \log_{10} 10 \)
\( = 2 \times 1 \)
\( = 2 \)
Thus, the limit is indeed 2.
In simple words: To prove this, we replace 'x' with '2' in the entire math problem. Then, we do all the calculations inside the log, which simplifies to 100. Since 100 is 10 squared, and the log is base 10, the answer is 2.

🎯 Exam Tip: When evaluating limits of composite functions like \( \log(f(x)) \), if \( \text{Lim}_{x \to a} f(x) \) exists and is positive, and the logarithm function is continuous, you can evaluate \( \log (\text{Lim}_{x \to a} f(x)) \). Also, remember common logarithm properties like \( \log_a a = 1 \) and \( \log_a a^b = b \).

Question 10. Given that \( f(x) = \frac{4-7x}{3x+4} \), \( l = \text{Lim}_{x \to 2} f(x) \) and \( m = \text{Lim}_{x \to 0} f(x) \), form the equation whose roots are \( \frac{1}{l}, \frac{1}{m} \).
Answer: First, we calculate the values of \( l \) and \( m \).
\( \implies l = \text{Lt}_{x \to 2} \frac{4-7x}{3x+4} \)
Substitute \( x = 2 \):
\( l = \frac{4-7 \times 2}{3 \times 2+4} = \frac{4-14}{6+4} = \frac{-10}{10} = -1 \)
\( \implies m = \text{Lt}_{x \to 0} \frac{4-7x}{3x+4} \)
Substitute \( x = 0 \):
\( m = \frac{4-7 \times 0}{3 \times 0+4} = \frac{4-0}{0+4} = \frac{4}{4} = 1 \)
Now we find the roots of the quadratic equation, which are \( \frac{1}{l} \) and \( \frac{1}{m} \).
First root \( = \frac{1}{l} = \frac{1}{-1} = -1 \)
Second root \( = \frac{1}{m} = \frac{1}{1} = 1 \)
Next, we calculate the sum of the roots (S) and the product of the roots (P).
\( S = \text{Sum of roots} = -1 + 1 = 0 \)
\( P = \text{Product of roots} = (-1) \times (1) = -1 \)
A quadratic equation with roots \( \alpha \) and \( \beta \) is given by \( x^2 - (\alpha+\beta)x + \alpha\beta = 0 \), or \( x^2 - Sx + P = 0 \).
Substituting S and P:
\( x^2 - (0)x + (-1) = 0 \)
\( x^2 - 1 = 0 \)
This is the required quadratic equation.
In simple words: First, find the value of 'l' by putting x=2 into the fraction, which gives -1. Then, find 'm' by putting x=0 into the fraction, which gives 1. The roots for our new equation are 1/l and 1/m, so they are -1 and 1. Add these roots to find the sum (S=0), and multiply them to find the product (P=-1). Finally, use the formula \( x^2 - Sx + P = 0 \) to get the equation \( x^2 - 1 = 0 \).

🎯 Exam Tip: To form a quadratic equation from its roots, always calculate the sum of the roots and the product of the roots first. Then, use the standard form \( x^2 - (\text{Sum of Roots})x + (\text{Product of Roots}) = 0 \). Be careful with signs when calculating sums and products, especially with negative values.