### Limits

• Understand concepts of limits- Algebraic, trigonometric, involving infinity, using L-Hôpital's Rule.

A limit is the value that a function or sequence "approaches" as the input or index approaches some value. Sometimes, a function is undefined if it takes some exact value of ’x’. In that case, we can find out where the function was approaching. This can be done by removing the factors causing undefined nature or sometimes by using specific mathematical identities.

## L Hôpital’s rule:

If a limit is in indeterminate form (such as $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ ) then,

$\lim_{x\to a}\dfrac{f(x)}{g(x)} = \dfrac{\lim_{x\to a}f'(x)}{\lim_{x\to a}g'(x)} = \dfrac{\lim_{x\to a}f''(x)}{\lim_{x\to a}g''(x)}$

### Solved Example:

#### 2-1-01

If a function is continuous at a point, then:

Solution:
A function f(x) is said to be continuous if the limit $\lim_{x\to a} f(x)$ exists and it is matching the actual value of the function irrespective of in what direction (from left or from right) we approach the point along the function. $\lim_{x\to a^-} f(x) = \lim_{x\to a^+} f(x) = f(a)$

Correct Answer: D

### Solved Example:

#### 2-1-02

Let x denote a real number. Find out the INCORRECT statement.

Solution:

S = {x :x $\in$ A AND x $\in$ B} represents the INTERSECTION of set A and set B.
S = {x :x $\in$ A OR x $\in$ B} represents the UNION of set A and set B.

Correct Answer: C

### Solved Example:

#### 2-1-03

Consider the function f(x) = $\mid x\mid$ in the interval -1 $\leq$ x $\leq$ 1. At the point x = 0, f (x) is:

Solution:

At x= 0, function is continuous but not differentiable because.

For x > 0, f'(x)=1, $\lim_{x\to 0^+}$ f'(x)=1

For x < 0, f'(x)=-1, $\lim_{x\to 0^-}$ f'(x)= -1

$\lim_{x\to 0^+}$ f'(x) $\neq$ $\lim_{x\to 0^-}$ f'(x)

Therefore, it is not differentiable.

Correct Answer: C

### Solved Example:

#### 2-1-04

$\lim_{x\to 0}\left( \dfrac {1-\cos x}{x^{2}}\right)$ is:

Solution:
$y=\lim _{x\to 0}\dfrac {\left( 1-\cos x\right) }{x^{2}}$ It forms 0/0 condition. Hence by L'Hopital rule, $y=\lim _{x\to 0}\dfrac {\dfrac {d}{dx}\left( 1-\cos x\right) }{\dfrac {d}{dx}x^{2}}$ Still this gives $\dfrac{0}{0}$ condition, so again applying L'Hopital rule, $y =\lim _{x\to 0}\dfrac {\dfrac {d}{dx}\left( \sin x \right) }{\dfrac {d}{dx}\left( 2x\right)} =\lim _{x\to 0}\dfrac {\cos x}{2} =\dfrac{\cos 0}{2} = \dfrac{1}{2}$

Correct Answer: B

### Solved Example:

#### 2-1-05

What is $\lim_{\theta \to 0}\dfrac {\sin \theta }{\theta}$ equal to?

Solution:
This is a standard identity or you may apply L'Hopital's rule, $y =\lim_{\theta \to 0}\dfrac {\sin \theta }{\theta } =\lim_{\theta \to 0}\dfrac {\dfrac {d}{d\theta }\sin \theta }{\dfrac {d}{d\theta }\theta } =\lim_{\theta \to 0}\dfrac {\cos \theta }{1} =\dfrac {\cos 0}{1} =1$

Correct Answer: D

### Solved Example:

#### 2-1-06

$\lim _{x\rightarrow 0} \dfrac{\sin^2x}{x}$ is equal to:

Solution:
\begin{align*} f\left( x\right) &=\lim _{x\rightarrow 0}\dfrac {\sin ^{2}x}{x}\\ &=\lim _{x\rightarrow 0}\dfrac {\sin ^{2}x}{x^{2}}\times \dfrac {x^{2}}{x}\\ &=\lim _{x\rightarrow 0}\left( \dfrac {\sin x}{x}\right) ^{2}\times x\\ &=\left( 1\right) ^{2}\times 0\\ &=0 \end{align*} Alternative method is to use the L'Hospital's Rule. Take derivative of both numerator as well as denominator. \begin{align*} f\left( x\right) &=\lim _{x\rightarrow 0}\dfrac {\sin ^{2}x}{x}f\left( x\right)\\ &=\lim _{x\rightarrow 0}\dfrac {2\sin x\cos x}{1}\\ &=\lim _{x\rightarrow 0}\dfrac {\sin 2x}{1}=\dfrac {\sin 0}{1}\\ &=0 \end{align*}

Correct Answer: A

### Solved Example:

#### 2-1-07

Which of the following functions is not differentiable in the domain [-1,1]?

Solution:

It clearly shows that for the first three functions, f (x) is differentiable at x =- 1, x = 0 and x = 1, i.e. in the domain [- 1, 1]. So, (a), (b) and (c) are differentiable.

Correct Answer: D

### Solved Example:

#### 2-1-08

At x = 0, the function f (x) = x$^3$+ 1 has:

Solution:

\begin{align*} f(x) &= x^3 + 1\\ f'(x) &= 3x^2,\quad f'(0) = 0\\ f''(x) &= 6x, \quad f''(0) = 0 \end{align*} f''(0) = 0 point of inflection.
f''(0) > 0 point of minima.
f''(0) < 0 point of maxima.

Correct Answer: D

### Solved Example:

#### 2-1-09

The minimum value of function y = $x^2$ in the interval [1, 5] is:

Solution:
Given :
y = x$^2$ ...(i) and interval [1, 5]
At x = 1 and y = 1,
And at x = 5, y = (5)$^2$ = 25
Here the interval is bounded between 1 and 5. So, the minimum value at this interval is 1.

Correct Answer: B

### Solved Example:

#### 2-1-10

The value of: $\lim _{x\rightarrow 8}\dfrac {x^{\frac{1}{3}}-2}{x-8}$

Solution:
Using the formula for, $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$
Here, a = x$^{\frac{1}{3}}$
and b = 2 \begin{align*} \lim _{x\rightarrow 8}\frac {x^{\frac{1}{3}}-2}{x-8} &= \dfrac {(x^{\frac{1}{3}}-2)}{(x^{\frac{1}{3}}-2)(x^{\frac{2}{3}} + {x^{\frac{1}{3}}(2) +4)}}\\ &= \dfrac {1}{(4)+ 2(2)+4}\\ &= \dfrac {1}{12} \end{align*}

Correct Answer: B

### Solved Example:

#### 2-1-11

If $f(x) = \dfrac{2x^2 -7x +3}{5x^2 -12x -9}$ then $\mathrm{lim}_{x\rightarrow 3} f(x)$ will be:

Solution:
First we will factorize numerator and denominator separately. \begin{align*} 2x^2 -7x +3 &=2x^{2}-6x-x+3\\ &=2x\left( x-3\right) -1\left( x-3\right)\\ &=\left( x-3\right) \left( 2x-1\right)\\ \end{align*} \begin{align*} 5x^{2}-12x-9 &=5x^{2}-15x+3x-9\\ &=5x\left( x-3\right) +3\left( x-3\right)\\ &=\left( x-3\right) \left( 5x+3\right) \end{align*} After cancelling the common factor from numerator and denominator, $f\left( x\right) =\dfrac {2x-1}{5x+3}$
Now substitute the value of x = 3, $f(x)=\dfrac {5}{18}$

Correct Answer: B

### Solved Example:

#### 2-1-12

$\lim_{x \to 1}(1-x)\tan \left( \dfrac{\pi x}{2} \right)$=?

Solution:
$\mathrm{Let\ } L = \lim_{x \to 1} \left( 1-x\right) \tan \left( \dfrac {\pi x}{2}\right)$ $\mathrm{Put}\ 1-x = \theta, \quad x=1-\theta, \quad \mathrm{As\ } x \rightarrow 1,\ \theta \rightarrow 0$ \begin{align*} L &= \lim_{\theta \to 0} \theta \tan \left[\dfrac {\pi }{2}\left( 1-\theta \right)\right] \\ &= \lim_{\theta \to 0}\theta \tan \left( \dfrac {\pi }{2}-\dfrac {\pi }{2}\theta \right)\\ &= \lim_{\theta \to 0}\theta \left[ \cot \left( \dfrac {\pi }{2}\theta \right) \right]\\ &= \lim_{\theta \to 0}\dfrac {\theta }{\tan \left( \dfrac {\pi }{2}\theta \right) }\\ &=\lim _{\theta \rightarrow 0}\dfrac {{\theta }}{\left[ \dfrac {\tan \left( \dfrac {\pi }{2}\theta \right) }{\left(\dfrac {\pi }{2}\theta \right) }\right]\cdot \dfrac {\pi }{2}{\theta }} =\dfrac{1}{1 \cdot \left(\dfrac {\pi }{2}\right)} =\dfrac {2}{\pi} \end{align*}

Correct Answer: C

### Solved Example:

#### 2-1-13

Let f: R $\rightarrow$ R be defined by: $(3x^2 +4) \cos x$ Then $\lim_{h \to 0} \dfrac{f(h) + f(-h) -8}{h^2}$ is equal to :

Solution:
$f\left( h\right) =\left( 3h^{2}+4\right) \cos\ h,$ $f\left( -h\right) =\left[ 3\left( -h\right) ^{2}+4\right] \cos \left( -h\right) =\left( 3h^{2}+4\right) \cos\ h$ Adding, $f\left( h\right) +f\left( -h\right) =\left( 6h^{2}+8\right) \cos\ h$
Subtract 8 from both sides, $f\left( h\right) +f\left( -h\right) -8 =6h^{2}\cos\ h +8\cos\ h -8$ \begin{align*} & \lim_{h \to 0} \dfrac {f\left( h\right) +f\left( -h\right) -8}{h^{2}}\\ &= \lim_{h \to 0} \dfrac{6h^{2}\cos\ h +8\cos\ h -8}{h^2}\\ &= \lim_{h \to 0} \dfrac {6h^{2}\cos\ h }{h^{2}}+8 \left(\dfrac {\cos\ h -1}{h^{2}}\right)\\ &= \lim_{h \to 0}6\cos\ h + 8\dfrac {\left(- 2\sin ^{2}\dfrac {h}{2}\right) }{h^{2}}\\ &=\lim_{h \to 0} 6\cos\ h - 16\left[ \dfrac{\left(\sin ^{2}\dfrac {h}{2}\right)}{\left( \dfrac {h}{2}\right) ^{2} \times 4}\right]\\ &=\lim_{h \to 0} 6\cos\ h -4 \lim_{\dfrac{h}{2} \to 0} \left[\dfrac {\sin ^{2}\left( \dfrac {h}{2}\right) }{\left( \dfrac {h}{2}\right) ^{2}}\right]\\ &=6(1)-4(1) =2 \end{align*}

Correct Answer: B

### Solved Example:

#### 2-1-14

The value of $\lim_{x \to \infty} \dfrac{x \ln x}{1 + x^2}$ is: (GATE Civil 2021)

Solution:
$\mathrm{Let\ L =} \lim_{x \to \infty} \dfrac{x \ln x}{1 + x^2}$ Applying L'Hopital's rule and taking derivatives of both numerator and denominator: $= \lim_{x \to \infty} \left[\dfrac{x\left(\dfrac{1}{x}\right) + \ln x}{2x} \right]$ Still this limit is in $\dfrac{\infty}{\infty}$ form, hence applying L'Hopital's rule one more time \begin{align*} L &= \lim_{x \to \infty} \left[\dfrac{0 + \dfrac{1}{x}}{2} \right]\\ &= \lim_{x \to \infty} \left[\dfrac{1}{2x} \right]\\ &= \dfrac{1}{2 \times \infty}\\ &= 0 \end{align*}

Correct Answer: D