CBSE Class 11 Mathematics Limits and Derivatives Notes Set 01

Important Notes and Formulas

Introduction

Calculus is the mathematics of motion and change, while algebra, geometry, and trigonometry are more static in nature. The development of calculus in the 17^th century by Newton, Leibnitz and others grew out of attempts by these and earlier mathematicians to answer certain fundamental questions about dynamic real-world situations. These investigations led to two fundamental procedures – differentiation and integration; which can be formulated in terms of a concept called limit.
In a very real sense, the concept of limit is the threshold to modern mathematics. You are about to cross that threshold, and beyond lies the fascinating world of calculus.

Limit of a Function (Informal Definition)

The notation \( \lim_{x \to c} f(x) = L \) is read "the limit of f(x) as x approaches c is L" and means that the functional values f(x) can be made arbitrarily close to a unique number L by choosing x sufficiently close to c (but not equal to c).

One-Sided Limits

(1) Right-hand limit: We write \( \lim_{x \to c^+} f(x) = L \) if we can make the number f(x) as close to L as we please by choosing x sufficiently close to c on a small interval (c, b) immediately to the right of c.
(2) Left-hand limit: We write \( \lim_{x \to c^-} f(x) = L \) if we can make the number f(x) as close to L as we please by choosing x sufficiently close to c on a small interval (a, c) immediately to the left of c.
(3) Limit of a function f(x) is said to exist as x → a (x approaches a) when \( \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = \ell \) (finite quantity)
⇒ \( \lim_{x \to a} f(x) = \ell \)

Evaluation of Algebraic Limits

Let f(x) be an algebraic function and 'a' be a real number. Then \( \lim_{x \to a} f(x) \) is known as an algebraic limit.
E.g. \( \lim_{x \to 1} \frac{x^3 - 1}{x - 1} \), \( \lim_{x \to 0} \frac{\sqrt{1 + x} - \sqrt{1 - x}}{x} \), \( \lim_{x \to 2} \frac{x^5 - 32}{x - 2} \) etc. are algebraic limits.

Direct Substitution Method

If by direct substitution of the point in the given expression we get a finite number, then the number obtained is the limit of the given expression.
If upon substituting the point in the given expression, we get the following forms:
\( \frac{0}{0}, \frac{\infty}{\infty} \), ∞ − ∞, 0 × ∞, ∞⁰, 0⁰, 1^∞ (Indeterminate Forms)
then we can't find the value of limit by direct substitution. Following methods are followed to find the limit of the function.

Factorisation Method

Consider \( \lim_{x \to a} \frac{f(x)}{g(x)} \). If by putting x = a the rational function \( \frac{f(x)}{g(x)} \) takes the form \( \frac{0}{0}, \frac{\infty}{\infty} \) etc., then (x − a) is a factor of both f(x) and g(x). In such a case we factorise the numerator and denominator, and then cancel out the common factor (x − a). After cancelling out the common factor (x − a), we again put x = a in the given expression and see whether we get a meaningful number or not. This process is repeated till we get a meaningful number.

Rationalisation Method

This is particularly used when either numerator or denominator, or both involve expressions consisting of square roots (radical signs).

Method of Evaluating Algebraic Limits when x → ∞

To evaluate this type of limits we follow the following procedure.
Step 1: Write down the given expression in the form of a rational function, i.e., \( \frac{f(x)}{g(x)} \), if it is not so.
Step 2: If k is the highest power of x in numerator and denominator both, then divide each term in numerator and denominator by x^k.
Step 3: Use the result \( \lim_{x \to \infty} \frac{1}{x^n} = 0 \), where n > 0.
Important Result: If m, n are positive integers and a₀, b₀ ≠ 0 are non-zero numbers, then
\( \lim_{x \to \infty} \frac{a_0 x^m + a_1 x^{m-1} + \ldots + a_{m-1} x + a_m}{b_0 x^n + b_1 x^{n-1} + \ldots + b_{n-1} x + b_n} = \begin{cases} \frac{a_0}{b_0}, & \text{if } m = n \\ 0, & \text{if } m < n \\ \infty, & \text{if } m > n,\ (a_0 \times b_0) > 0 \\ -\infty, & \text{if } m > n,\ (a_0 \times b_0) < 0 \end{cases} \)

Standard Algebraic Limit

\( \lim_{x \to a} \left( \frac{x^n - a^n}{x - a} \right) = n \cdot a^{n-1} \)

Trigonometric Limits

To evaluate trigonometric limits the following results are very useful.
1. \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \)
2. \( \lim_{x \to 0} \frac{\tan x}{x} = 1 \)
(in 1 and 2, x is measured in radians)
3. \( \lim_{x \to 0} \frac{\sin x^\circ}{x} = \frac{\pi}{180} \)
4. \( \lim_{x \to 0} \cos x = 1 \)
5. \( \lim_{x \to a} \frac{\sin(x - a)}{x - a} = 1 \)
6. \( \lim_{x \to a} \frac{\tan(x - a)}{x - a} = 1 \)

The Algebra of Limits

Let f and g be two real functions with domain D. We define four new functions f ± g, fg, f/g on domain D by setting
(f ± g)(x) = f(x) ± g(x), (fg)(x) = f(x) × g(x), (f/g)(x) = f(x)/g(x), if g(x) ≠ 0 for any x ∈ D.
Following are some results concerning the limits of these functions.
Let \( \lim_{x \to a} f(x) = \ell \) and \( \lim_{x \to a} g(x) = m \), if ℓ and m exist.
1. \( \lim_{x \to a} (f \pm g)(x) = \lim_{x \to a} f(x) \pm \lim_{x \to a} g(x) = \ell \pm m \)
2. \( \lim_{x \to a} (fg)(x) = \lim_{x \to a} f(x) \times \lim_{x \to a} g(x) = \ell m \)
3. \( \lim_{x \to a} \left( \frac{f}{g} \right)(x) = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)} = \frac{\ell}{m} \), provided m ≠ 0.
4. \( \lim_{x \to a} k f(x) = k \cdot \lim_{x \to a} f(x) \), where k is constant.
5. If \( \lim_{x \to a} f(x) = +\infty \) or −∞, then \( \lim_{x \to a} \frac{1}{f(x)} = 0 \).

Solved Examples

Example 1. Evaluate \( \lim_{x \to 2} \frac{x^3 - 6x^2 + 11x - 6}{x^2 - 6x + 8} \).
Answer:
When x = 2, the expression \( \frac{x^3 - 6x^2 + 11x - 6}{x^2 - 6x + 8} \) is of the form \( \frac{0}{0} \).
Now, \( \lim_{x \to 2} \frac{x^3 - 6x^2 + 11x - 6}{x^2 - 6x + 8} \) \( \left( \frac{0}{0} \text{ form} \right) \)
\( = \lim_{x \to 2} \frac{(x - 1)(x - 2)(x - 3)}{(x - 2)(x - 4)} \)
\( = \lim_{x \to 2} \frac{(x - 1)(x - 3)}{(x - 4)} = \frac{(2 - 1)(2 - 3)}{(2 - 4)} = \frac{1}{2} \)

Example 2. Evaluate \( \lim_{x \to a} \frac{\sqrt{a + 2x} - \sqrt{3x}}{\sqrt{3a + x} - 2\sqrt{x}} \).
Answer:
When x = a, \( \frac{\sqrt{a + 2x} - \sqrt{3x}}{\sqrt{3a + x} - 2\sqrt{x}} \) assumes the indeterminate form \( \frac{0}{0} \).
Now, \( \lim_{x \to a} \frac{\sqrt{a + 2x} - \sqrt{3x}}{\sqrt{3a + x} - 2\sqrt{x}} = \lim_{x \to a} \frac{(\sqrt{a + 2x} - \sqrt{3x})(\sqrt{a + 2x} + \sqrt{3x})(\sqrt{3a + x} + 2\sqrt{x})}{(\sqrt{3a + x} - 2\sqrt{x})(\sqrt{3a + x} + 2\sqrt{x})(\sqrt{a + 2x} + \sqrt{3x})} \)
\( = \lim_{x \to a} \frac{(a - x)(\sqrt{3a + x} + 2\sqrt{x})}{3(a - x)(\sqrt{a + 2x} + \sqrt{3x})} \)
\( = \lim_{x \to a} \frac{\sqrt{3a + x} + 2\sqrt{x}}{3(\sqrt{a + 2x} + \sqrt{3x})} = \frac{4\sqrt{a}}{3 \cdot 2\sqrt{3a}} = \frac{2}{3\sqrt{3}} \)

Example 3. Evaluate the following limit: \( \lim_{x \to \frac{\pi}{2}} \frac{\sin(\cos x) \cos x}{\sin x - \operatorname{cosec} x} \) (using \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \))
Answer:
\( \lim_{x \to \frac{\pi}{2}} \frac{\sin(\cos x) \cos x}{\sin x - \operatorname{cosec} x} = \lim_{x \to \frac{\pi}{2}} \frac{\sin(\cos x) \sin x \cos x}{\sin^2 x - 1} \)
\( = \lim_{x \to \frac{\pi}{2}} \frac{\sin(\cos x) \sin x \cos x}{-\cos^2 x} = -\lim_{x \to \frac{\pi}{2}} \frac{\sin(\cos x)}{\cos x} \cdot \sin x \)
As \( x \to \frac{\pi}{2} \), cos x → 0, so \( \frac{\sin(\cos x)}{\cos x} \to 1 \) and sin x → 1.
= −1.

Example 4. Solve: \( \lim_{x \to \pi/3} \frac{\tan^3 x - 3\tan x}{\cos\left(x + \frac{\pi}{6}\right)} \).
Answer:
Let \( L = \lim_{x \to \pi/3} \frac{\tan^3 x - 3\tan x}{\cos\left(x + \frac{\pi}{6}\right)} \) and \( x - \frac{\pi}{3} = t \)
Since \( \tan 3x = \frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x} \), we have \( \tan^3 x - 3\tan x = \tan 3x \left( 3\tan^2 x - 1 \right) \).
⇒ \( L = \lim_{t \to 0} \frac{\tan(3t + \pi)\left[3\tan^2\left(t + \frac{\pi}{3}\right) - 1\right]}{\cos\left(t + \frac{\pi}{2}\right)} = \lim_{t \to 0} \frac{\tan(3t)\left[3\tan^2\left(t + \frac{\pi}{3}\right) - 1\right]}{-\sin t} \)
⇒ \( L = -3 \lim_{t \to 0} \frac{\tan(3t)}{3t} \times \lim_{t \to 0} \frac{t}{\sin t} \times \lim_{t \to 0} \left[3\tan^2\left(t + \frac{\pi}{3}\right) - 1\right] \)
⇒ L = −3 × 1 × 1 × (3 × 3 − 1) = −3 × 8 = −24.

Example 5. Evaluate the following limit: \( \lim_{x \to 2} \frac{3^x + 3^{3 - x} - 12}{3^{3 - x} - 3^{x/2}} \).
Answer:
\( L = \lim_{x \to 2} \frac{3^x + 3^{3 - x} - 12}{3^{3 - x} - 3^{x/2}} \)
⇒ \( L = \lim_{x \to 2} \frac{3^x + \frac{27}{3^x} - 12}{\frac{27}{3^x} - 3^{x/2}} \)
⇒ \( L = \lim_{x \to 2} \frac{3^{2x} - 12 \cdot 3^x + 27}{-\left(3^{3x/2} - 3^3\right)} \)
⇒ \( L = -\lim_{x \to 2} \frac{(3^x - 9)(3^x - 3)}{\left(3^{x/2} - 3\right)\left(3^x + 9 + 3 \cdot 3^{x/2}\right)} \)
⇒ \( L = -\lim_{x \to 2} \frac{\left(3^{x/2} + 3\right)(3^x - 3)}{3^x + 3 \cdot 3^{x/2} + 9} \)
⇒ \( L = -\frac{6 \cdot 6}{9 + 3 \cdot 3 + 9} = -\frac{36}{27} = -\frac{4}{3} \).

Example 6. Evaluate the left hand and right hand limits of the function defined by \( f(x) = \begin{cases} 1 + x^2, & \text{if } 0 \le x \le 1 \\ 2 - x, & \text{if } x > 1 \end{cases} \) at x = 1. Also, show that \( \lim_{x \to 1} f(x) \) does not exist.
Answer:
We have, (LHL of f(x) at x = 1) \( = \lim_{x \to 1^-} f(x) = \lim_{h \to 0} f(1 - h) \)
\( = \lim_{h \to 0} 1 + (1 - h)^2 = \lim_{h \to 0} 2 - 2h + h^2 = 2 \)
and, (RHL of f(x) at x = 1) \( = \lim_{x \to 1^+} f(x) = \lim_{h \to 0} f(1 + h) \)
\( = \lim_{h \to 0} 2 - (1 + h) = \lim_{h \to 0} (1 - h) = 1 \)
Clearly, \( \lim_{x \to 1^-} f(x) \ne \lim_{x \to 1^+} f(x) \). So, \( \lim_{x \to 1} f(x) \) does not exist.

Example 7. If \( f(x) = \begin{cases} \frac{x - |x|}{x}, & x \ne 0 \\ 2, & x = 0 \end{cases} \) show that \( \lim_{x \to 0} f(x) \) does not exist.
Answer:
We have, (LHL of f(x) at x = 0) \( = \lim_{x \to 0^-} f(x) = \lim_{h \to 0} f(0 - h) = \lim_{h \to 0} \frac{-h - |-h|}{(-h)} \)
\( = \lim_{h \to 0} \frac{-h - h}{-h} = \lim_{h \to 0} \frac{-2h}{-h} = \lim_{h \to 0} 2 = 2 \)
(RHL of f(x) at x = 0) \( = \lim_{x \to 0^+} f(x) = \lim_{h \to 0} f(0 + h) = \lim_{h \to 0} \frac{h - |h|}{h} \)
\( = \lim_{h \to 0} \frac{h - h}{h} = \lim_{h \to 0} \frac{0}{h} = \lim_{h \to 0} 0 = 0 \)
Clearly, \( \lim_{x \to 0^-} f(x) \ne \lim_{x \to 0^+} f(x) \). So, \( \lim_{x \to 0} f(x) \) does not exist.

Example 8. Discuss the existence of each of the following limits: (i) \( \lim_{x \to 0} \frac{1}{x} \) (ii) \( \lim_{x \to 0} \frac{1}{|x|} \)
Answer:
(i) The graph of \( f(x) = \frac{1}{x} \) is as shown in the figure below. We observe that as x approaches 0 from the LHS i.e. x is negative and very close to zero, then the values of \( \frac{1}{x} \) are negative and very large in magnitude.
∴ \( \lim_{x \to 0^-} \frac{1}{x} \to -\infty \)

Similarly, when x approaches 0 from the right i.e. x is positive and very close to 0, then the values of \( \frac{1}{x} \) are very large and positive.
∴ \( \lim_{x \to 0^+} \frac{1}{x} \to \infty \). Thus we have, \( \lim_{x \to 0^-} \frac{1}{x} \ne \lim_{x \to 0^+} \frac{1}{x} \)
Hence, \( \lim_{x \to 0} \frac{1}{x} \) does not exist.
(ii) The graph of \( f(x) = \frac{1}{|x|} \) is shown in the figure below. We observe that as x approaches 0 from LHS i.e. x is negative and close to 0, then |x| is close to zero and is positive. Consequently, \( \frac{1}{|x|} \) is large and positive.
∴ \( \lim_{x \to 0^-} \frac{1}{|x|} \to \infty \)

Also, if x approaches 0 from RHS i.e. x is positive and close to 0, then |x| is close to zero and is positive. Consequently, \( \frac{1}{|x|} \) is large and positive.
∴ \( \lim_{x \to 0^+} \frac{1}{|x|} \to \infty \)
Thus, \( \frac{1}{|x|} \) becomes large and positive from both sides of 0, and we write \( \lim_{x \to 0} \frac{1}{|x|} = \infty \). Since ∞ is not a real number, the limit is not finite (it does not exist as a real number); the function simply grows without bound.

Example 9. If \( f(x) = \begin{cases} mx^2 + n, & x < 0 \\ nx + m, & 0 \le x \le 1 \\ nx^3 + m, & x > 1 \end{cases} \) For what values of integers m, n do the limits \( \lim_{x \to 0} f(x) \) and \( \lim_{x \to 1} f(x) \) exist?
Answer:
It is given that \( \lim_{x \to 0} f(x) \) and \( \lim_{x \to 1} f(x) \) both exist
⇔ \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) \) and \( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) \)
⇔ \( \lim_{h \to 0} f(0 - h) = \lim_{h \to 0} f(0 + h) \) and \( \lim_{h \to 0} f(1 - h) = \lim_{h \to 0} f(1 + h) \)
⇔ \( \lim_{h \to 0} m(-h)^2 + n = \lim_{h \to 0} n(h) + m \) and \( \lim_{h \to 0} n(1 - h) + m = \lim_{h \to 0} n(1 + h)^3 + m \)
⇔ n = m and n + m = n + m
⇔ m = n
Hence, \( \lim_{x \to 0} f(x) \) and \( \lim_{x \to 1} f(x) \) both exist for m = n.

Example 10. If \( f(x) = \begin{cases} |x| + 1, & x < 0 \\ 0, & x = 0 \\ |x| - 1, & x > 0 \end{cases} \). For what value(s) of a does \( \lim_{x \to a} f(x) \) exist?
Answer:
We have, \( f(x) = \begin{cases} -x + 1, & x < 0 \\ 0, & x = 0 \\ x - 1, & x > 0 \end{cases} \) \( \left[ \because |x| = \begin{cases} x, & x \ge 0 \\ -x, & x < 0 \end{cases} \right] \)
Clearly, \( \lim_{x \to a} f(x) \) exists for all a ≠ 0. So, let us see whether \( \lim_{x \to 0} f(x) \) exists or not.
We have, \( \lim_{x \to 0^-} f(x) = \lim_{h \to 0} f(0 - h) = \lim_{h \to 0} -(-h) + 1 = 1 \)
\( \lim_{x \to 0^+} f(x) = \lim_{h \to 0} f(0 + h) = \lim_{h \to 0} h - 1 = -1 \)
∴ \( \lim_{x \to 0^-} f(x) \ne \lim_{x \to 0^+} f(x) \)
So, \( \lim_{x \to 0} f(x) \) does not exist. Hence, \( \lim_{x \to a} f(x) \) exists for all a ≠ 0.

Example 11. Suppose \( f(x) = \begin{cases} a + bx, & x < 1 \\ 4, & x = 1 \\ b - ax, & x > 1 \end{cases} \) and if \( \lim_{x \to 1} f(x) = f(1) \). What are possible values of a and b?
Answer:
We have, \( \lim_{x \to 1} f(x) = f(1) \)
⇔ \( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) \)
⇔ \( \lim_{x \to 1^-} f(x) = f(1) \) and \( \lim_{x \to 1^+} f(x) = f(1) \)
⇔ \( \lim_{h \to 0} f(1 - h) = 4 \) and \( \lim_{h \to 0} f(1 + h) = 4 \)
⇔ \( \lim_{h \to 0} \{a + b(1 - h)\} = 4 \) and \( \lim_{h \to 0} \{b - a(1 + h)\} = 4 \)
⇔ a + b = 4 and b − a = 4
⇔ a = 0, b = 4

Example 12. Evaluate the left hand and right hand limits of the function \( f(x) = \begin{cases} \frac{\sqrt{x^2 - 6x + 9}}{(x - 3)}, & x \ne 3 \\ 0, & x = 3 \end{cases} \) at x = 3 and hence comment on the existence of limit at x = 3.
Answer:
The given function can be written as \( f(x) = \begin{cases} \frac{|x - 3|}{(x - 3)}, & x \ne 3 \\ 0, & x = 3 \end{cases} \)
∴ L.H.L. \( = \lim_{x \to 3^-} f(x) = \lim_{h \to 0} f(3 - h) = \lim_{h \to 0} \frac{|3 - h - 3|}{(3 - h - 3)} = \lim_{h \to 0} \frac{|-h|}{(-h)} = \lim_{h \to 0} \frac{h}{-h} = -1 \)
and R.H.L. \( = \lim_{x \to 3^+} f(x) = \lim_{h \to 0} f(3 + h) = \lim_{h \to 0} \frac{|3 + h - 3|}{(3 + h - 3)} = \lim_{h \to 0} \frac{|h|}{h} = \lim_{h \to 0} \frac{h}{h} = 1 \)
Hence left hand limit and right hand limit of f(x) at x = 3 are −1 and 1 respectively.
As Left Hand Limit ≠ Right Hand Limit, i.e. \( \lim_{x \to 3} f(x) \) does not exist.