and 
are fixed non-zero constants and 
and 
are integers) : 
.Assume, $f(x) = \frac{x}{{{{\sin }^n}x}}$. Upon differentiating with respect to $x$ and applying quotient rule we get, ${f^\prime }(x) = \frac{{{{\sin }^n}x\frac{d}{{dx}}x - x\frac{d}{{dx}}{{\sin...
and are fixed non-zero constants and and are integers) : 
.Assume, $f(x) = (x + \sec x)(x - \tan x)$. Upon differentiating with respect to $x$ and applying quotient rule we get, ${f^\prime }(x) = (x + \sec x)\frac{d}{{dx}}(x - \tan x) + (x - \tan...
and are fixed non-zero constants and and are integers) : 
.Assume, $f(x) = \frac{x}{{1 + \tan x}}$. Upon differentiating with respect to $x$ and applying quotient rule we get, ${f^\prime }(x) = \frac{{(1 + \tan x)\frac{d}{{dx}}(x) - x\frac{d}{{dx}}(1 + \tan...
and are fixed non-zero constants and and are integers) : 
.Assume, $f(x) = \frac{{{x^2}\cos \left( {\frac{\pi }{4}} \right)}}{{\sin x}}$. Upon differentiating with respect to $x$ and applying quotient rule we get, ${f^\prime }(x) = \cos \frac{\pi }{4} \cdot...
and are fixed non-zero constants and and are integers) : 
.Assume, $f(x) = \frac{{4x + 5\sin x}}{{3x + 7\cos x}}$. Upon differentiating with respect to $x$ and applying quotient rule we get, ${f^\prime }(x) = \frac{{(3x + 7\cos x)\frac{d}{{dx}}(4x + 5\sin...
and are fixed non-zero constants and and are integers) : 
.Assume, $f(x) = (x + \cos x)(x - \tan x)$ Upon differentiating with respect to $x$ and applying quotient rule we get, ${f^\prime }(x) = (x + \cos x)\frac{d}{{dx}}(x - \tan x) + (x - \tan...
and are fixed non-zero constants and and are integers): 
.Assume, $f(x) = \left( {a{x^2} + \sin x} \right)(p + q\cos x)$. Upon differentiating with respect to $x$ and applying quotient rule we get, ${f^\prime }(x) = \left( {a{x^2} + \sin x}...
and are fixed non-zero constants and and are integers) : 
.Assume, $f(x) = \left( {{x^2} + 1} \right)\cos x$. Upon differentiating with respect to $x$ and applying quotient rule we get, ${f^\prime }(x) = \left( {{x^2} + 1} \right)\frac{d}{{dx}}(\cos x) +...
and are fixed non-zero constants and and are integers): 
.Assume, $f(x) = {x^4}(5\sin x - 3\cos x)$. Upon differentiating with respect to $x$ and applying quotient rule we get, ${f^\prime }(x) = {x^4}\frac{d}{{dx}}(5\sin x - 3\cos x) + (5\sin x - 3\cos...
and are fixed non-zero constants and and are integers): 
.Assume, $f(x) = \frac{{\sin (x + a)}}{{\cos x}}$. Upon differentiating with respect to $x$ and applying quotient rule we get, ${f^\prime }(x) = \frac{{\cos x\frac{d}{{dx}}[\sin (x + a)] - \sin (x +...
and are fixed non-zero constants and and are integers): 
.Assume, $f(x) = \frac{{a + b\sin x}}{{c + d\cos x}}$ Upon differentiating with respect to $x$ and applying quotient rule we get, $f'(x) = \frac{{(c + d\cos x)\frac{d}{{dx}}(a + b\sin x) - (a + b\sin...
and are fixed non-zero constants and and are integers): 
.Assume, $y = {\sin ^n}x$ When $n = 1$, then $y = \sin x$ Differentiating with respect to $x$ we get, $\frac{{dy}}{{dx}} = \cos x$ This implies, $\frac{d}{{dx}}\sin x = \cos x$. Now when $n = 2$,...
and are fixed non-zero constants and and are integers): 
.Suppose $f(x) = \frac{{\sec x - 1}}{{\sec x + 1}}$ Simplifying the given function to get, $f(x) = \frac{{\frac{1}{{\cos x}} - 1}}{{\frac{1}{{\cos x}} + 1}}$ $f(x) = \frac{{1 - \cos x}}{{1 + \cos...
and are fixed non-zero constants and and are integers): 
.Suppose $f(x) = \frac{{\sin x + \cos x}}{{\sin x - \cos x}}$ Upon differentiating with respect to $x$ and applying quotient rule we get, ${f^\prime }(x) = \frac{{(\sin x - \cos x)\frac{d}{{dx}}(\sin...
, where 
.We are given, $f(x) = \left| x \right| - 5$ Evaluating $\mathop {\lim }\limits_{x \to 5} f(x)$ When, $\mathop {\lim }\limits_{x \to {5^ - }} f(x) = \mathop {\lim }\limits_{x \to {5^ - }} |x| - 5$ $...
, where 
.We are given, $f(x)=\left\{ \begin{array}{*{35}{l}} \frac{x}{|x|},x\ne 0 \\ 0,x=0 \\ \end{array} \right.$ Evaluating $\mathop {\lim }\limits_{x \to 0} f(x)$, When, $\mathop {\lim }\limits_{x \to...
, where 
.We are given, $f(x) = \left\{ \begin{gathered} {x^2} - 1,x \leqslant 1 \hfill \\ - {x^2} - 1,x > 1 \hfill \\ \end{gathered} \right.$ Evaluating $\underset{x\to 1}{\mathop{\lim }}\,f(x)$,...
and , where 
.We have the function, $f(x) = \left\{ \begin{gathered} 2x + 3x \leqslant 0 \hfill \\ 3\left( {x + 1} \right)x > 0 \hfill \\ \end{gathered} \right.$ Evaluating $\mathop {\lim }\limits_{x \to 0}...
.Given, $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\tan 2x}}{{x - \frac{\pi }{2}}}$ Substituting $x = \frac{\pi }{2}$ to get, $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\tan...
.Given, $\mathop {\lim }\limits_{x \to 0} (\operatorname{cosec} x - \cot x)$ Using trigonometric ratios for, $\operatorname{cosec} x = \frac{1}{{\sin x}}$ and $\cot x = \frac{{\cos x}}{{\sin x}}$ to...
.Given, $\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax + bx}}{{ax + \sin bx}}$ Substituting $x = 0$ to get, $\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax + bx}}{{ax + \sin bx}} = \frac{0}{0}$...
.Given, $\mathop {\lim }\limits_{x \to 0} x\sec x$ $\mathop {\lim }\limits_{{\text{x}} \to 0} \operatorname{xsec} {\text{x}} = \mathop {\lim }\limits_{{\text{x}} \to 0} \frac{{\text{x}}}{{\cos...
.Given, $\mathop {\lim }\limits_{x \to 0} \frac{{ax + x\cos x}}{{b\sin x}}$ $\mathop {\lim }\limits_{x \to 0} \frac{{ax + x\cos x}}{{b\sin x}} = \frac{0}{0}$ As, this limit is undefined so the other...
.We are given, $\mathop {\lim }\limits_{x \to 0} \frac{{\cos 2x - 1}}{{\cos x - 1}}$ Substituting $x = 0$ to get, $\mathop {\lim }\limits_{x \to 0} \frac{{\cos 2x - 1}}{{\cos x - 1}} = \frac{0}{0}$...
.We are given, $\mathop {\lim }\limits_{x \to 0} \frac{{\cos x}}{{\pi - x}}$ Substituting $x = 0$ to get, $\mathop {\lim }\limits_{x \to 0} \frac{{\cos x}}{{\pi - x}} = \frac{{\cos 0}}{{\pi - 0}}$...
(vii) Assume, $f(x) = 2\tan x - 7\sec x$. Using the first principle we get, ${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}$ ${f^\prime }(x) = \mathop {\lim...
(vi) 
(v) Assume, $f(x) = 3\cot x + 5\operatorname{cosec} x$. Upon differentiating with respect to $x$ to get, ${{\text{f}}^\prime }({\text{x}}) = 3{(\cot x)^\prime } + 5{(\operatorname{cosec}...
(iv) 
(iii) Assume, $f(x) = 5\sec x + 4\cos x$. Upon differentiating with respect to $x$ to get, ${{\text{f}}^\prime }({\text{x}}) = \frac{{\text{d}}}{{{\text{dx}}}}(5\sec {\text{x}} + 4\cos {\text{x}})$...
at 
.Assume, $f(x) = 99x$, By the first principle we have, ${{\text{f}}^\prime }({\text{x}}) = \mathop {\lim }\limits_{{\text{h}} \to 0} \frac{{{\text{f}}({\text{x}} + {\text{h}}) -...
(ii) 
.(i) Assume, $f{\text{ }}\left( x \right){\text{ }} = {\text{ }}{x^3}\;--{\text{ }}27$. Using first principle we have, ${{\text{f}}^\prime }({\text{x}}) = \mathop {\lim }\limits_{{\text{h}} \to 0}...
(iv) 
(iii) Assume, $\;f\left( x \right) = \frac{1}{{{x^2}}}$. Using first principle we have, ${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}$ $f'(x) = \mathop {\lim...
.Prove that 
.We are given, $f(x) = \frac{{{x^{100}}}}{{100}} + \frac{{{x^{99}}}}{{99}} + \ldots \frac{{{x^2}}}{2} + x + 1$ Differentiating with respect to $x$ to get, $\frac{d}{{dx}}f(x) = \frac{d}{{dx}}\left[...
for some fixed real number 
.We are given, ${x^n} + a{x^{n - 1}} + {a^2}{x^{n - 2}} + \ldots + {a^{n - 1}}x + {a^n}$. Differentiating with respect to $x$ to get, ${f^\prime }(x) = \frac{d}{{dx}}\left( {{x^n} + a{x^{n - 1}} +...
(ii) 
.(i) Assume, ${\text{f(x) = (x - a)(x - b)}}$ ${\text{f}}({\text{x}}) = {{\text{x}}^2} - ({\text{a}} + {\text{b}}){\text{x}} + {\text{ab}}$ Differentiating with respect to $x$ to get,...
(iii) Assume, $f(x) = \frac{{(x - a)}}{{(x - b)}}$. Upon differentiating both sides and using quotient rule to get, ${f^\prime }(x) = \frac{d}{{dx}}\left( {\frac{{x - a}}{{x - b}}} \right)$...
for some constant .Assume, $f(x) = \frac{{{x^n} - {a^n}}}{{x - a}}$. Upon differentiating both sides and using quotient rule to get, ${f^\prime }(x) = \frac{d}{{dx}}\left( {\frac{{{x^n} - {a^n}}}{{x - a}}} \right)$...
(ii) 
.(i) Assume, $f(x) = \frac{{2x - 3}}{4}$. Differentiating with respect to $x$ to get, ${f^\prime }(x) = \frac{d}{{dx}}\left( {2x - \frac{3}{4}} \right)$ ${f^\prime }(x) = 2\frac{d}{{dx}}(x) -...
(iv) 
(iii) Assume, $f(x) = {x^{ - 3}}(5 + 3x)$. Upon differentiating with respect to $x$ and applying Leibnitz product rule to get, ${f^\prime }(x) = {x^{ - 3}}\frac{d}{{dx}}(5 + 3x) + (5 +...
(vi) 
(v) Assume, $f(x) = {x^{ - 4}}\left( {3 - 4{x^{ - 5}}} \right)$. Upon differentiating with respect to $x$ and applying Leibnitz product rule to get, ${f^\prime }(x) = {x^{ - 4}}\frac{d}{{dx}}\left(...
(ii) 
.(i) Assume, $f(x) = \sin x\cos x$. Using the first principle we get, ${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}$ $f'(x) = \mathop {\lim }\limits_{h \to 0}...
from first principle.Assume, $f(x) = \cos x$. Then, $f(x + h) = \cos (x + h)$. Using first principle we get, ${{\text{f}}^\prime }({\text{x}}) = \mathop {\lim }\limits_{h \to 0} \frac{{{\text{f}}({\text{x}} +...
and are fixed non-zero constants and and are integers): 
.Suppose $f(x) = \frac{{\cos x}}{{1 + \sin x}}$ By applying quotient rule we get, ${f^\prime }(x) = \frac{{(1 + \sin x)\frac{d}{{dx}}(\cos x) - (\cos x)\frac{d}{{dx}}(1 + \sin x)}}{{{{(1 + \sin...
at 
.Suppose $f(x) = x$. By the first principle we have, ${{\text{f}}^\prime }({\text{x}}) = \mathop {\lim }\limits_{{\text{h}} \to 0} \frac{{{\text{f}}({\text{x}} + {\text{h}}) -...
at 
.Suppose $f(x) = {x^2} - 2$. By the first principle we have, ${{\text{f}}^\prime }({\text{x}}) = \mathop {\lim }\limits_{{\text{h}} \to 0} \frac{{{\text{f}}({\text{x}} + {\text{h}}) -...
and are fixed non-zero constants and and are integers): 
.Suppose $f(x) = \operatorname{cosec} x\cot x$. By Leibnitz product rule we have, ${f^\prime }(x) = \operatorname{cosec} x{(\cot x)^\prime } + \cot x{(\operatorname{cosec} x)^\prime...
and are fixed non-zero constants and and are integers): 
Suppose $f(x) = \sin (x + a)$ This implies $f(x + h) = \sin (x + h + a)$ By the first principle we have, ${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}$ $ = \mathop...
and are fixed non-zero constants and and are integers): 
Suppose $f(x) = {(ax + b)^n}{(cx + d)^{rn}}$ By Leibnitz product rule we have, ${f^\prime }(x) = {(ax + b)^n}\frac{d}{{dx}}{(cx + d)^{\prime \prime \prime }} + {(cx + d)^m}\frac{d}{{dx}}{(ax +...
and are fixed non-zero constants and and are integers): 
Suppose $f(x) = {(ax + b)^n}$ This implies $f(x + h) = {\{ a(x + h) + b\} ^N} = {(ax + ah + b)^n}$ By the first principle we have, ${f^\prime }(x) = \mathop {\lim }\limits_{s \to 0} \frac{{f(x + h)...
and are fixed non-zero constants and and are integers): 
Suppose $f(x) = 4\sqrt x - 2$ Upon differentiating with respect to $x$ we get, ${f^\prime }(x) = \frac{d}{{dx}}(4\sqrt x - 2) = \frac{d}{{dx}}(4\sqrt x ) - \frac{d}{{dx}}(2)$ $ =...
and are fixed non-zero constants and and are integers): 
Suppose $f(x) = \frac{a}{{{x^4}}} - \frac{b}{{{x^2}}} + \cos x$ Upon differentiating with respect to $x$ we get, ${f^\prime }(x) = \frac{d}{{dx}}\left( {\frac{a}{{{x^4}}}} \right) -...
and are fixed non-zero constants and and are integers): 
Suppose $f(x) = \frac{{p{x^2} + qx + r}}{{ax + b}}$ By applying quotient rule we get, ${f^\prime }(x) = \frac{{(ax + b)\frac{d}{{dx}}\left( {p{x^2} + qx + r} \right) - \left( {p{x^2} + qx + r}...
and are fixed non-zero constants and and are integers): 
Suppose $f(x) = \frac{{ax + b}}{{p{x^2} + qx + r}}$ By applying quotient rule we get, ${f^\prime }(x) = \frac{{\left( {p{x^2} + qx + r} \right)\frac{d}{{dx}}(ax + b) - (ax + b)\frac{d}{{dx}}\left(...
and are fixed non-zero constants and and are integers): 
Suppose $f(x) = \frac{1}{{a{x^2} + bx + c}}$ By applying quotient rule we get, ${f^\prime }(x) = \frac{{\left( {a{x^2} + bx + c} \right)\frac{d}{{dx}}(1) - \frac{d}{{dx}}\left( {a{x^2} + bx + c}...
and are fixed non-zero constants and and are integers): ![\[\frac{{1 + \frac{1}{x}}}{{1 - \frac{1}{x}}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1ed4de8079f93d8d01ad1dfd6f4379f7_l3.png "Rendered by QuickLaTeX.com")
Suppose, \[f(x) = \frac{{1 + \frac{1}{x}}}{{1 - \frac{1}{x}}}\] \[f(x) = \frac{{\frac{{x + 1}}{x}}}{{\frac{{x - 1}}{x}}}\] \[f(x) = \frac{{x + 1}}{{x - 1}}\] By applying quotient rule we get,...
and are fixed non-zero constants and and are integers): 
.Suppose $f(x) = \frac{{ax + b}}{{cx + d}}$ By applying quotient rule we get, \[{f^\prime }(x) = \frac{{(cx + d)\frac{d}{{dx}}(ax + b) - (ax + b)\frac{d}{{dx}}(cx + d)}}{{{{(cx + d)}^2}}}\] $f'(x) =...
and are fixed non-zero constants and and are integers): 
Suppose $f(x) = (ax + b){(cx + d)^2}$ By Leibnitz product rule we have, ${f^\prime }(x) = (ax + b)\frac{d}{{dx}}{(cx + d)^2} + {(cx + d)^2}\frac{d}{{dx}}(ax + b)$ $ = (ax + b)\frac{d}{{dx}}\left(...
and are fixed non-zero constants and and are integers): 
Suppose $f(x) = (px + q)\left( {\frac{r}{x} + s} \right)$ By Leibnitz product rule we have, \[{f^\prime }(x) = (px + q){\left( {\frac{r}{x} + s} \right)^\prime } + \left( {\frac{r}{x} + s}...
and are fixed non-zero constants and and are integers): 
Suppose $f(x) = x + a$ This implies \[{\text{f(x + h) = x + h + a}}\] By the first principle we have, ${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}$ $ = \mathop...
(iv) 
(iii) Suppose, $f(x) = \sin (x + 1)$ This implies, $f(x + h) = \sin (x + h + 1)$ By the first principle we have, ${{\text{f}}^\prime }({\text{x}}) = \mathop {\lim }\limits_{h \to 0}...
(ii) 
.( i ) Suppose, $f(x) = - x$. This implies, $f(x + h) = - (x + h)$ By the first principle we have, ${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}$ $ = \mathop {\lim...
, evaluate .We are given that $f(x)$ satisfies, $\mathop {\lim }\limits_{x \to 1} \frac{{f(x) - 2}}{{{x^2} - 1}} = \pi $ $\frac{{\mathop {\lim }\limits_{x \to 1} {\text{f}}(x) - 2}}{{\mathop {\lim }\limits_{x...
be fixed real numbers and define a function 
. What is 
? For some 
, Compute 
.We are given, $f(x) = \left( {x - {a_1}} \right)\left( {x - {a_2}} \right)....\left( {x - {a_n}} \right)$ Evaluating $\mathop {\lim }\limits_{{\text{x}} \to {{\text{a}}_1}} {\text{f}}({\text{x}})$...
, where .We are given, $f(x) = \left| x \right| - 5$ Evaluating $\mathop {\lim }\limits_{x \to 5} f(x)$ When, $\mathop {\lim }\limits_{x \to {5^ - }} f(x) = \mathop {\lim }\limits_{x \to {5^ - }} |x| - 5$ \[...
, where 
.We are given, $f(x) = \left\{ \begin{gathered} \frac{x}{{\left| x \right|}},x \ne 0 \hfill \\ 0,x = 0 \hfill \\ \end{gathered} \right.$ Evaluating $\mathop {\lim }\limits_{x \to 0} f(x)$, When,...
, where 
.We are given, $f(x) = \left\{ \begin{gathered} \frac{{\left| x \right|}}{x},x \ne 0 \hfill \\ 0,x = 0 \hfill \\ \end{gathered} \right.$ For $\mathop {\lim }\limits_{x \to a} f(x)$ to exist, then...
, where 
.We are given, $f(x) = \left\{ \begin{gathered} {x^2} - 1,x \leqslant 1 \hfill \\ - {x^2} - 1,x > 1 \hfill \\ \end{gathered} \right.$ Evaluating $\underset{x\to 1}{\mathop{\lim }}\,f(x)$ , When,...
and , where 
.We have the function, $f(x) = \left\{ \begin{gathered} 2x + 3x \leqslant 0 \hfill \\ 3\left( {x + 1} \right)x > 0 \hfill \\ \end{gathered} \right.$ Evaluating $\mathop {\lim }\limits_{x \to 0}...
.Given, $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\tan 2x}}{{x - \frac{\pi }{2}}}$ Substituting $x = \frac{\pi }{2}$ to get, $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\tan...
.Given, $\mathop {\lim }\limits_{x \to 0} (\operatorname{cosec} x - \cot x)$ Using trigonometric ratios for, $\operatorname{cosec} x = \frac{1}{{\sin x}}$ and $\cot x = \frac{{\cos x}}{{\sin x}}$ to...
.Given, $\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax + bx}}{{ax + \sin bx}}$ Substituting $x = 0$ to get, $\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax + bx}}{{ax + \sin bx}} = \frac{0}{0}$...
.Given, $\mathop {\lim }\limits_{x \to 0} x\sec x$ $\mathop {\lim }\limits_{{\text{x}} \to 0} \operatorname{xsec} {\text{x}} = \mathop {\lim }\limits_{{\text{x}} \to 0} \frac{{\text{x}}}{{\cos...
.Given, $\mathop {\lim }\limits_{x \to 0} \frac{{ax + x\cos x}}{{b\sin x}}$ $\mathop {\lim }\limits_{x \to 0} \frac{{ax + x\cos x}}{{b\sin x}} = \frac{0}{0}$ As, this limit is undefined so the other...
.We are given, $\mathop {\lim }\limits_{x \to 0} \frac{{\cos 2x - 1}}{{\cos x - 1}}$ Substituting $x = 0$ to get, $\mathop {\lim }\limits_{x \to 0} \frac{{\cos 2x - 1}}{{\cos x - 1}} = \frac{0}{0}$...
.We are given, $\mathop {\lim }\limits_{x \to 0} \frac{{\cos x}}{{\pi - x}}$ Substituting $x = 0$ to get, $\mathop {\lim }\limits_{x \to 0} \frac{{\cos x}}{{\pi - x}} = \frac{{\cos 0}}{{\pi - 0}}$...
.We are given, $\mathop {\lim }\limits_{x \to \pi } \frac{{\sin (\pi - x)}}{{\pi (\pi - x)}}$ $\mathop {\lim }\limits_{x \to \pi } \frac{{\sin (\pi - x)}}{{\pi (\pi - x)}} = \mathop {\lim...
We are given, $\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{{\sin bx}},a,b \ne 0$ Substituting $x = 0$ to get, $\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{{\sin bx}} = \frac{0}{0}$ As,...
.We are given, $\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{{bx}}$ Substituting $x = 0$ to get, $\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{{bx}} = \frac{0}{0}$ As, this limit is...
.Given, $\mathop {\lim }\limits_{x \to - 2} \frac{{\frac{1}{x} + \frac{1}{2}}}{{x + 2}}$ Substituting $x = - 2$ to get, $\mathop {\lim }\limits_{x \to - 2} \frac{{\frac{1}{x} + \frac{1}{2}}}{{x +...
.We are given, $\mathop {\lim }\limits_{x \to 1} \frac{{a{x^2} + bx + c}}{{c{x^2} + bx + a}},a + b + c \ne 0$ Now, substituting $x = 1$ to get, $\mathop {\lim }\limits_{x \to 1} \frac{{a{x^2} + bx +...
.We are given, $\mathop {\lim }\limits_{z \to 1} \frac{{{z^{\frac{1}{3}}} - 1}}{{{z^{\frac{1}{6}}} - 1}}$ Substituting $z = 1$ to get, $\mathop {\lim }\limits_{z \to 1} \frac{{{z^{\frac{1}{3}}} -...
.We are given, $\mathop {\lim }\limits_{x \to 0} \frac{{ax + b}}{{cx + 1}}$ Substituting $x = 0$to get, $\mathop {\lim }\limits_{x \to 0} \frac{{ax + b}}{{cx + 1}} = \frac{{a(0) + b}}{{c(0) + 1}}$ $...
.We are given, $\mathop {\lim }\limits_{x \to 3} \frac{{{x^4} - 81}}{{2{x^2} - 5x - 3}}$ Substituting the limit at $x = 3$ to get, $\mathop {\lim }\limits_{x \to 3} \frac{{{x^4} - 81}}{{2{x^2} - 5x -...
.We are given, $\mathop {\lim }\limits_{x \to 2} \frac{{3{x^2} - x - 10}}{{{x^2} - 4}}$ Substituting the limit at $x = 2$ to get, $\mathop {\lim }\limits_{x \to 2} \frac{{3{x^2} - x - 10}}{{{x^2} -...
.We are given the limit, $\mathop {\lim }\limits_{x \to 0} \frac{{{{(x + 1)}^5} - 1}}{x}$ Substituting the value $x = 0$ to get, $ = \frac{{{{\left( {0 + 1} \right)}^5} - 1}}{0}$ As, this limit is...
.We have to evaluate, $\mathop {\lim }\limits_{x \to - 1} \frac{{{x^{10}} + {x^5} + 1}}{{x - 1}}$ Substituting the value $x = - 1$ to get, $\mathop {\lim }\limits_{x \to - 1} \frac{{{x^{10}} +...
.We have to evaluate, $\mathop {\lim }\limits_{x \to 4} \frac{{4x + 3}}{{x - 2}}$ Substituting the value $x = 4$ to get, $\mathop {\lim }\limits_{x \to 4} \frac{{4x + 3}}{{x - 2}} = \frac{{\left[...
.We have to evaluate, $\mathop {\lim }\limits_{r \to 1} \pi {r^2}$ Substituting the value $r = 1$to get, $\mathop {\lim }\limits_{r \to 1} \pi {r^2} = \pi {(1)^2}$ $ = \pi $.
.We have to evaluate, $\mathop {\lim }\limits_{x \to n} \left( {x - \frac{{22}}{7}} \right)$ Substituting the value $x = \pi $ to get, $ = \pi - \frac{{22}}{7}$.
.We have to evaluate, $\mathop {\lim }\limits_{x \to 3} x + 3$ Substituting the value $x=3$ to get, $ = 3 + 3$ $ = 6$.