Basics of Limits

Find the limit.

$\underset{x\to 2}{\lim }\left(\frac{\sqrt{1-\cos \left\{2\left(x-2\right)\right\}}}{x-2}\right)$

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

It is given that $\underset{x\to 0}{\lim }\frac{\frac{x^2}{{\left(a+x^r\right)}^{{\displaystyle \frac{1}{p}}}}}{1-\cos \mathit{x}}=\ell$. If $P=3$ and $\ell =1$, then the value of $a$ is equal to

If $\text{f}\left(\text{n, }\theta \right)=\underset{\text{r}=1}{\overset{\text{n}}{\Pi }}\left(1-{\text{tan}}^2\frac{\theta }{2^{\text{r}}}\right)$, then compute $\underset{\text{n}\to \infty }{\text{Lim}}\text{ f }\left(\text{n}, \theta \right)$       { given $\theta \ne 0$ }

Let $f\mathit{: }R\mathit{\to }R$ be such that $f\left(1\right)=3$ and $f^{\text{'}}\left(1\right)=6$. Then $\underset{x\to 0}{\mathrm{Lim}}{\left(\frac{f\left(1+x\right)}{f\left(1\right)}\right)}^{1/x}$

If $\underset{x\to 0}{\lim }\frac{\sin \left(nx\right)[(a-n)nx-\tan x]}{x^2}=0, (n>0)$ then the value of $\text{'}a\text{'}$ is equal to

Evaluate: $\underset{x\to \infty }{\lim }{\left(x\right)}^{e^{-x}}$.

Evaluate: $\underset{x\to 0}{\lim }{x}^{\tan x}$

Evaluate: $\underset{x\to -\infty }{\lim }x{e}^x$.

$\underset{x\to 1}{\lim }\left(\left(1-x\right)+\left[x-1\right]+\left|1-x\right|\right)$ 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,

The value of$\underset{n\to \infty }{\lim }{\sin }^{40}(n!x\pi ),$ where $n\in N$ and $x$ is a rational number, is:

$\underset{x\to 0}{\lim }\frac{e^{1/x}}{e^{\left(1/x\right)+1}}$   is equal to

$\underset{x\to 0}{\mathrm{Lim}}\frac{1-{\cos }^{3}x}{x.\sin x.\cos x}$ equals

$\underset{x\to 0}{\lim }\frac{\left(1-\cos 2x\right)\left(3+\cos x\right)}{x\tan 4x}$ is equal to

$\underset{x\to \frac{\pi }{6}}{\lim }\frac{\sin 2x}{\sin x}$ is equal to

If  $f\left(x\right)=\left\{\begin{array}{l}\frac{\sin \left[x\right]}{\left[x\right]},\left[x\right]\ne 0\\ 0 , \left[x\right]=0\end{array}\right.$, where $[.]$ denotes the greatest integer function, then find $\underset{x\to 0}{\lim }f\left(x\right)$.

Let $l_n=\frac{2^n+{\left(-2\right)}^n}{2^n}$ and $L_n=\frac{2^n+{\left(-2\right)}^n}{3^n}$ then as $n\to \infty$

$\underset{x\to 0}{\lim }\frac{\tan \left(\left[-{\pi }^2\right]x^2\right)-x^2\tan \left[-{\pi }^2\right]}{{\sin }^{2}x}$ denotes the greatest integers function