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...

We are given, $f(x) = \left( {x - {a_1}} \right)\left( {x - {a_2}} \right)....\left( {x - {a_n}} \right)$ Evaluating $\mathop {\lim }\limits_{{\text{x}} \to {{\text{a}}_1}} {\text{f}}({\text{x}})$...

We are given, $f(x) = \left| x \right| - 5$ Evaluating $\mathop {\lim }\limits_{x \to 5} f(x)$ When, $\mathop {\lim }\limits_{x \to {5^ - }} f(x) = \mathop {\lim }\limits_{x \to {5^ - }} |x| - 5$ \[...

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,...

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...

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,...

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$.