Basics of Limits

If $\underset{x\to \infty }{\lim }\frac{\left(e^{\mu x}+5\right)}{\left(e^{100x}+7\right)}$ exists, then sum of all possible positive integral values of $\mu$ is

Evaluate the limit $\underset{x\to 0}{\lim }\left(\frac{2x}{\left|x\right|+x^2}\right)$.

If $f\left(x\right)=8x^3+3x$, then $\underset{x\to \infty }{\lim }\left[\frac{f^{-1}\left(8x\right)-f^{-1}\left(x\right)}{x^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}\right]$ is

Let $$\begin{gathered} f\left(x\right)=\left\{\begin{array}{l}x; x<1 \\ 3; x=1 \\ 2-x^2; 1<x\le 2\\ x-3; x>2\end{array}\right. \end{gathered}$$

Evaluate each of the following limits, if exists:

$\left(\mathrm{i}\right) \underset{x\to 1^{-}}{\lim }f\left(x\right)$

$\left(\mathrm{ii}\right) \underset{x\to 1^{+}}{\lim }f\left(x\right)$

$\left(\mathrm{iii}\right) \underset{x\to 1}{\lim }f\left(x\right)$

$\left(\mathrm{iv}\right) \underset{x\to 2^{-}}{\lim }f\left(x\right)$

$\left(\mathrm{v}\right) \underset{x\to 2^{+}}{\lim }f\left(x\right)$

$\left(\mathrm{vi}\right) \underset{x\to 2}{\lim }f\left(x\right)$

$\underset{n\to \infty }{\lim }\left[\frac{n{\left(1^3+2^3+3^3+....+n^3\right)}^2}{{\left(1^2+2^2+3^2+...+n^2\right)}^3}\right]=$

Evaluate: $\underset{n\to \infty }{\lim }\left(\frac{1^3+2^3+3^3+...+n^3}{\sqrt{4n^8+1}}\right)$

Find $\underset{n\to \infty }{\lim }\left(1+\frac{1}{a_1}\right)\left(1+\frac{1}{a_2}\right)\left(1+\frac{1}{a_3}\right)......\left(1+\frac{1}{a_n}\right)$ where $a_1=1$ and $a_n=n\left(1+a_{n-1}\right); n\ge 2$.

$\underset{n\to \infty }{\lim }{\left({}^{2n}C_n\right)}^{\frac{1}{n}}=$

Solve: $\underset{x\to 0}{\lim }\left(\frac{{\left|x\right|}^{\alpha }}{e^x}\right); \alpha \in R^{+}$

Let the variable $x_n$ be determined by the following law of formation

$x_0=\sqrt{a}$, $x_1=\sqrt{a+\sqrt{a}}$, $x_2=\sqrt{a+\sqrt{a+\sqrt{a}}}$, $x_3=\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a}}}}$.

Find $\underset{n\to \infty }{\lim }{x}_n$.

$\underset{x\to \infty }{\lim }\left[\frac{1}{1-x^2}+\frac{2}{1-x^2}+\frac{3}{1-x^2}+...+\frac{x}{1-x^2}\right]$ is:

Evaluate $\underset{n\to \infty }{\lim }\left(n\sin \left(2\mathrm{\pi }\sqrt{1+n^2}\right)\right); n\in N$.

Function whose jump (non-negative difference of LHL & RHL) of discontinuity is greater than or equal to one, is/are -

$\underset{x\to \infty }{\lim }\sqrt{\frac{x-\sin \left(\cos \left(\tan x\right)\right)}{4x+\left(\cos \left(\sin x\right)\right)}}$ equals

For $x\to 0$ determine the order of smallness, relative to the infinitesimal $\beta \left(x\right)=x$, of the following infinitesimals:

$\frac{x^2\left(1+x\right)}{1+\sqrt[3]{x}}$

$\underset{x\to 0}{\lim }{x}^{\frac{3}{2}}\left(x>0\right)=$

If $A_i=\frac{x-a_i}{\left|x-a_i\right|},i=1,2,3,\dots .,n$ and $a_1<a_2<a_3<\dots ..<a_n$, $\underset{x\to a_m}{\lim }\left(A_1{A}_2\dots .A_n\right),1\le m\le n$

(b) Plot the graph of the function $f\left(x\right)=\underset{t\to 0}{\mathrm{Lim}} \left(\frac{2x}{\pi }{\tan }^{-1}\frac{x}{t^2}\right)$

$\underset{x\to 1}{\lim }\left(\left(1-x\right)+\left[x-1\right]+\left|1-x\right|\right)$ where $\left[x\right]$ denotes the greatest integer less than or equal to $x$

Let $f\left(x\right)=x\left[\frac{1}{x}\right]$ for $x(\ne 0)\in R,$ where for each $t\in R, \left[t\right]$ denotes the greatest integer less than or equal to $t.$ Then,