Limits of Exponential and Logarithmic Functions

Show that $\underset{x\to 0}{\lim } \frac{e^{\tan x}-e^x}{\tan x-x}=1$

$\underset{x\to 0}{\lim }\left(\left(\frac{1-{\cos }^2\left(3x\right)}{{\cos }^3\left(4x\right)}\right)\left(\frac{{\sin }^3\left(4x\right)}{{\left({\log }_e\left(2x+1\right)\right)}^5}\right)\right)$ is equal to

Evaluate $\underset{x\to 0}{\lim }\left(\frac{1}{x}\ln \sqrt{\frac{1+x}{1-x}}\right)$

$\underset{x\to 0}{\lim }\frac{2^{2x}-2^{x+1}+2-\cos 2x}{x^2}=$

Find the value of limit $\underset{n\to \infty }{\lim }\left({\left(\frac{n}{n}\right)}^n+{\left(\frac{n-1}{n}\right)}^n+{\left(\frac{n-2}{n}\right)}^n+..........{\left(\frac{1}{n}\right)}^n\right)$ 

$\underset{h\to 0}{\lim }\frac{e^{{\left(x+h\right)}^2}-e^{x^2}}{h}=$

Find the limit of $\underset{x\to 0}{\lim }\left(\frac{e^x-1}{\sqrt{1+x}-1}\right).$

$\underset{x\to 0}{\lim }\frac{\ln \left(1+x^2+x^4\right)}{x\left(e^x-1\right)}=$

Find the limit $\underset{x\to 0}{\lim }{\left(\frac{\sin x}{x}\right)}^{\frac{\sin x}{x-\sin x}}.$

Find the limit $\underset{x\to 0}{\lim }{\left(\cos x\right)}^{\frac{1}{\sin x}}.$

Let $\mathit{\text{f}}\text{: }\mathit{\text{R}}\to \mathit{\text{R}}$ be such that $\mathit{\text{f}}\left(1\right)=2\text{, }{\mathit{\text{f}}}^{\text{'}}\left(1\right)=4$, then find the $\underset{\mathit{\text{x}}\to \mathrm{0 }}{\text{lim}}{\left[\frac{\mathit{\text{f}}\left(1+\mathit{\text{x}}+{\mathit{\text{x}}}^2\right)+\mathit{\text{xf}}\left(1+\mathit{\text{x}}\right)}{\mathit{\text{f}}\left(1\right)}\right]}^{\frac{1}{\text{x}\left(\mathit{\text{x}}+1\right)}}$.

$\underset{h\to 0}{\lim }\frac{\log (1+2h)-2\log (1+h)}{h^2}=$_____

Evaluate:$\underset{x\to \mathrm{ }0 }{\mathrm{l}\mathrm{i}\mathrm{m} }\left(\frac{1}{x}\right)\mathrm{l}\mathrm{o}\mathrm{g}\left(\frac{2+x}{2-x}\right)$

Evaluate : $\underset{n\to \infty }{\lim }{\left(1+\sin \frac{a}{n}\right)}^n$ .

Find $\underset{x\to 0}{\lim }\frac{e^{x^2}-\cos x}{{\sin }^2 x}$ 

Let $f:R\to R$, be such that $f\left(1\right)=2, f^{\text{'}}\left(1\right)=4$, then $\underset{x\to 0 }{\lim }{\left[\frac{f\left(1+x+x^2\right)+xf\left(1+x\right)}{f\left(1\right)}\right]}^{\frac{1}{x\left(x+1\right)}}=\_\_\_\_\_\_$

If $\underset{\mathrm{x}\to 0^{+}}{\lim }\left\{\frac{p+e^{{\displaystyle \frac{1}{x}}}}{1+q{e}^{{\displaystyle \frac{1}{x}}}}\right\}=2$ then which is NOT possible.

If $\mathrm{\ell }=\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{asinx}+\mathrm{bcos}2\mathrm{x}-2-\mathrm{x}+4{\mathrm{x}}^2}{{\mathrm{x}}^2}$ exists and finite then which of the following option(s) is/are correct?

If $\underset{x\to 0}{\lim }\frac{a{e}^x+b\cos x+c{e}^{-x}}{e^{2x}-2e^x+1}$$=4$, then

The value of $\underset{x\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{\ln \left(1+2x+4x^2\right)+\ln \left(1-2x+4x^2\right)}{\sec x-\cos x}$ is equal to