Introduction to Limit of a Function and Indeterminate Forms

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

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

$\underset{\mathrm{x}\to \frac{\mathrm{\pi }}{6}}{lim}\frac{2-\sqrt{3}\cos \mathrm{x}+\sin \mathrm{x}}{(6\mathrm{x}-\mathrm{\pi }{)}^2}$

Evaluate $\underset{x\to 0}{\lim } {\left(1+mx\right)}^{\frac{m}{x}}$

If $f\left(x\right)=\ln x,$then $\underset{x\to 3}{\lim }\frac{f\left(x\right)-f\left(3\right)}{x-3}$ is 

If $f\left(x\right)=\ln x,$then $\underset{x\to 3}{\lim }\frac{f\left(x\right)-f\left(3\right)}{x-3}$ is 

Evaluate $\underset{x\to \infty }{\lim }{x}^{\frac{1}{x-4}}$.

Evaluate $\underset{x\to \infty }{\lim }{x}^{\frac{1}{x-3}}$.

Evaluate $\underset{x\to \infty }{\lim }{x}^{\frac{1}{x-2}}$.

Evaluate $\underset{x\to \infty }{\lim }{x}^{\frac{1}{x-1}}$.

Evaluate $\underset{x\to 4}{\lim }{\left(x-4\right)}^{\left(x-4\right)}$.

Evaluate $\underset{x\to 3}{\lim }{\left(x-3\right)}^{\left(x-3\right)}$.

Evaluate $\underset{x\to 2}{\lim }{\left(x-2\right)}^{\left(x-2\right)}$.

Evaluate $\underset{n\to \infty }{\lim }{2}^{n-1}\sin \left(\frac{a}{2^n}\right)$. [Write the answer in fraction.]

Evaluate $\underset{x\to \infty }{\lim } \left\{2^{x-1}\tan \left(\frac{a}{2^x}\right)\right\}$. [Write the answer in fraction.]

Evaluate $\underset{x\to \infty }{\lim } x\tan \left(\frac{1}{x}\right)$.

The value of $\underset{n\to \infty }{\lim } 2^n \sin \left(\frac{a}{2^n}\right)$ is:

Evaluate $\underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\left(\frac{\mathrm{\pi }}{2}-x\right)\tan x$

Evaluate $\underset{x\to 1}{\lim }{\left(x-1\right)}^{\left(x-1\right)}$.

Evaluate: $\underset{x\to 1}{\lim }\left(\frac{1}{x^2+x-2}-\frac{x}{x^3-1}\right)$. [Write the answer in fraction.]