9.1 Questions

1.1 Basic Method of Evaluation of Limits:
1.2 Questions
1.3 Formal definition of Limit:
1.4 Evaluation of Limits by Direct Substitution Method:
1.5 Neighbourhood Concept:

2.1 One - Sided Limits:
2.2 Left hand Limit of a function:(LHL)
2.3 Right hand Limit of a function:(RHL)

6.1 Factorization
6.2 Rationalization
6.3 Series Expansions:

7.1 Limits of the form ${1^\infty }$
7.2 Limits of the form ${0^0}$
7.3 Limits of the form ${\infty^0}$
7.4 Limit of a function as $x \to \infty $ :

9.1 Questions

Evaluate $\mathop {\lim }\limits_{x \to 0} \frac{{2\sin (x) - \sin (2x)}}{{x - \sin (x)}}$

As $x \to 0$, $\frac{{2\sin (x) - \sin (2x)}}{{x - \sin (x)}} \to \frac{0}{0}$

$\mathop {\lim }\limits_{x \to 0} \frac{{2\sin (x) - \sin (2x)}}{{x - \sin (x)}}$ $ = \mathop {\lim }\limits_{x \to 0} \frac{{2\cos (x) - 2\cos (2x)}}{{1 - \cos (x)}}$ [By using L'hospital's Rule]

$ = \mathop {\lim }\limits_{x \to 0} \frac{{ - 2\sin (x) + 4\sin (2x)}}{{\sin (x)}}$ [By using L'hospital's Rule]

$ = \mathop {\lim }\limits_{x \to 0} \frac{{ - 2\cos (x) + 8\cos (2x)}}{{\cos (x)}}$ [By using L'hospital's Rule]

$ = \frac{{ - 2 + 8}}{1}$

$ = 6$

L'Hospital rule has certain limitations :

1.Can be applied only when limit takes the form $\frac{0}{0}$ or $\frac{\infty }{\infty }$

2.After differentiating numerator and denominator for finite number of times, we need to obtain a function whose limit can be obtained by using standard methods of evaluation of Limits

Evaluate $\mathop {\lim }\limits_{x \to \infty } \frac{x}{{{{({x^2} + 1)}^{0.5}}}}$

If we try to apply L'Hospital Rule, we will repeatedly end up with functions whose limits are unknown to us.

so,

$\mathop {\lim }\limits_{x \to \infty } \frac{x}{{{{({x^2} + 1)}^{0.5}}}} = \mathop {\lim }\limits_{x \to \infty } \frac{x}{{x{{\left( {1 + \frac{1}{{{x^2}}}} \right)}^{0.5}}}} = \mathop {\lim }\limits_{x \to \infty } \frac{1}{{{{\left( {1 + \frac{1}{{{x^2}}}} \right)}^{0.5}}}} = 1$

Evaluate $\mathop {\lim }\limits_{u \to - 2} \frac{{\sin \left( {\pi u} \right)}}{{{u^2} - 4}}$.

As ${u \to - 2}$, $\frac{{\sin \left( {\pi u} \right)}}{{{u^2} - 4}} \to \frac{0}{0}$

so, we can apply L'Hospital Rule.

$\mathop {\lim }\limits_{u \to - 2} \frac{{\sin \left( {\pi u} \right)}}{{{u^2} - 4}} = \mathop {\lim }\limits_{u \to - 2} \frac{{\pi \cos (\pi u)}}{{2u}}$

=$\frac{{\pi \cos (\pi ( - 2))}}{{2( - 2)}}$

$ = \frac{\pi }{{ - 4}}$

L'Hospital can be applied only when limit takes the form $\frac{0}{0}$ or $\frac{\infty }{\infty }$. so if given function is not in the form of $\frac{0}{0}$ or $\frac{\infty }{\infty }$, to apply L'Hospital rule, it must be rearranged in such a fashion that it takes the form of $\frac{0}{0}$ or $\frac{\infty }{\infty }$

Evaluate $\mathop {\lim }\limits_{x \to {0^ + }} \sqrt x \cdot \ln x$.

As ${x \to {0^ + }}$, $\sqrt x \cdot \ln x$ takes the form $0 \cdot \infty $.

So L'Hospital rule won't work.However, we can turn this into a fraction if we rewrite things as following

${x \to {0^ + }}$, $\sqrt x \cdot \ln x$ takes the form $0 \cdot \infty $ = $\mathop {\lim }\limits_{x \to {0^ + }} \frac{{\ln x}}{{\frac{1}{{\sqrt x }}}}$

and the limit is now in the form $ - \frac{\infty }{\infty }$ and we can now use L’Hospital’s Rule.

$\mathop {\lim }\limits_{x \to {0^ + }} \frac{{\ln x}}{{\frac{1}{{\sqrt x }}}} = \mathop {\lim }\limits_{x \to {0^ + }} \;\frac{{\frac{1}{x}}}{{\frac{{ - 1}}{{2({x^{3/2}})}}}}$

$ = \mathop {\lim }\limits_{x \to {0^ + }} - 2\sqrt x $

=0

Let $f(x) = \left\{ \begin{gathered} ax + 2{\text{ , x < 1}} \hspace{1em} \\ {{\text{x}}^2}{\text{ , x > 1}} \hspace{1em} \\ \end{gathered} \right.$

If $\mathop {\lim }\limits_{x \to 1} f(x)$ exists finitely.

Find the value of $a$.

RHL:

$\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} {x^2} = 1$

Given $\mathop {\lim }\limits_{x \to 1} f(x)$ exists finitely.

$ \Rightarrow $ LHL = RHL

$ \Rightarrow \mathop {\lim }\limits_{x \to {1^ - }} f(x) = 1$

$ \Rightarrow \mathop {\lim }\limits_{x \to {1^ - }} ax + 2 = 1$

$ \Rightarrow $ $a(1) + 2 = 1$

$ \Rightarrow a = - 1$

Evaluate $\mathop {\lim }\limits_{x \to {0^ + }} {\left( {\sin x} \right)^x}$.

As ${x \to {0^ + }}$, ${\left( {\sin x} \right)^x}$ takes the form of ${0^0}$.

So $\mathop {\lim }\limits_{x \to {0^ + }} {\left( {\sin x} \right)^x} = {e^{\mathop {\lim }\limits_{x \to {0^ + }} x\log \left( {\sin x} \right)}}$

$ = {e^{\mathop {\lim }\limits_{x \to {0^ + }} \frac{{\log (\sin x)}}{{\frac{1}{x}}}}}$

$ = {e^{\mathop {\lim }\limits_{x \to {0^ + }} \frac{{\frac{1}{{\sin x}} \cdot \cos x}}{{\frac{{ - 1}}{{{x^2}}}}}}}$ [applying L'Hospital's rule]

$ = {e^{ - \mathop {\lim }\limits_{x \to {0^ + }} {x^2}\cot x}}$

$ = {e^{ - \mathop {\lim }\limits_{x \to {0^ + }} \frac{{{x^2}}}{{\tan x}}}}$

$ = {e^0}$

$ = 1$

Evaluate $\mathop {\lim }\limits_{x \to 0} {\left( {cosec} \right)^x}$.

As ${x \to 0}$, ${\left( {cosec} \right)^x}$ takes the form ${\infty ^0}$

So, $\mathop {\lim }\limits_{x \to 0} {\left( {cosec} \right)^x} = {e^{\mathop {\lim }\limits_{x \to 0} x\log \left( {cosecx} \right)}}$

$ = {e^{\mathop {\lim }\limits_{x \to 0} \frac{{\log (cosecx)}}{{\frac{1}{x}}}}}$

$ = {e^{\mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{cosecx}} \cdot ( - cosecx\ \cot x)}}{{\frac{{ - 1}}{{{x^2}}}}}}}$ [applying L'Hospital rule]

$ = {e^{\mathop {\lim }\limits_{x \to 0} \frac{{{x^2}}}{{\tan x}}}}$

$ = {e^0}$

$ = 1$

If $f(x) = \sin x$. Evaluate $\mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}$.

$\mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{\sin (x + h) - \sin x}}{h}$

$ = \mathop {\lim }\limits_{h \to 0} \frac{{\sin x\cosh + \sinh \cos x - \sin x}}{h}$

$ = \mathop {\lim }\limits_{h \to 0} \frac{{\sin x(\cosh - 1) + \sinh \cos x}}{h}$

$ = \mathop {\lim }\limits_{h \to 0} \frac{{\sin x(\cosh - 1)}}{h} + \mathop {\lim }\limits_{h \to 0} \frac{{\sinh \cos x}}{h}$

$ = \sin x\mathop {\lim }\limits_{h \to 0} \frac{{(\cosh - 1)}}{h} + \cos x\mathop {\lim }\limits_{h \to 0} \frac{{\sinh }}{h}$

$ = \sin x\mathop {\lim }\limits_{h \to 0} \frac{{ - 2{{\sin }^2}\frac{h}{2}}}{h} + \cos x\mathop {\lim }\limits_{h \to 0} \frac{{\sinh }}{h}$

$ = - \sin x\mathop {\lim }\limits_{\frac{h}{2} \to 0} \frac{{\frac{h}{2}{{\sin }^2}\frac{h}{2}}}{{\frac{h}{2} \cdot \frac{h}{2}}} + \cos x\mathop {\lim }\limits_{h \to 0} \frac{{\sinh }}{h}$ [as ${h \to 0}$, ${\frac{h}{2} \to 0}$ ]

$ = \frac{{ - \sin x}}{2} \cdot 0 \cdot 1 + \cos x \cdot 1$ [using standard limit $\mathop {\lim }\limits_{j \to 0} \frac{{\sin j}}{j} = 1$ and $\mathop {\lim }\limits_{j \to 0} j = 0$ ]

$ = \cos x$

Let $f(x)$ be a function of $x$, then $\mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}$ is called as derivative of $f(x)$.And derivative of if $f(x)$ is denoted by $\frac{d}{{dx}}f(x)$

Hence, derivative of $sin(x) = cos(x)$.

i.e; $\frac{d}{{dx}}\sin x = \cos x$

Evaluate $\mathop {\lim }\limits_{n \to \infty } \frac{{[x] + [2x] + [3x] +..... + [nx]}}{{{n^2}}}$, where $[.]$ denotes the greatest integer function.

We know that $x - 1 < [x] \leqslant x$

$ \Rightarrow x + 2x +..... + nx - n < [x] + [2x] + [3x] +..... + [nx] \leqslant x + 2x +..... + nx$

$ \Rightarrow $ $\frac{{x \cdot n(n + 1)}}{2} - n < [x] + [2x] + [3x] +..... + [nx] \leqslant \frac{{x \cdot n(n + 1)}}{2}$

$ \Rightarrow $ $\frac{x}{2}\left( {1 + \frac{1}{n}} \right) - \frac{1}{n} < \frac{{[x] + [2x] + [3x] +..... + [nx]}}{{{n^2}}} \leqslant \frac{x}{2}\left( {1 + \frac{1}{n}} \right)$

$ \Rightarrow $ $\mathop {\lim }\limits_{n \to \infty } \frac{x}{2}\left( {1 + \frac{1}{n}} \right) - \frac{1}{n} = \frac{x}{2}$ and $\mathop {\lim }\limits_{n \to \infty } \frac{x}{2}\left( {1 + \frac{1}{n}} \right) = \frac{x}{2}$

$\therefore $ by sandwich theorem $\mathop {\lim }\limits_{n \to \infty } \frac{{[x] + [2x] + [3x] +..... + [nx]}}{{{n^2}}} = \frac{x}{2}$.

Evaluate $\mathop {\lim }\limits_{x \to 0} \frac{{\cos x - {{\cos }^3}x}}{{{x^2}}}$.

$\mathop {\lim }\limits_{x \to 0} \frac{{\cos x - {{\cos }^3}x}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\cos x(1 - {{\cos }^2}x)}}{{{x^2}}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{\cos x{{\sin }^2}x}}{{{x^2}}}$

$ = \mathop {\lim }\limits_{x \to 0} \cos x{\left( {\frac{{\sin x}}{x}} \right)^2}$

$ = 1$