Standard Limits

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 $\theta \ne 0$ }

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 $\underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\frac{{\left(1-\sin x\right)}^2}{{\left({\displaystyle \frac{\mathrm{\pi }}{2}-x}\right)}^2}$

Find the value of $\underset{x\to 0}{\lim }\frac{2\sin x-\sin 2x}{x^3}$

Define a sequence $\left\{S_n\right\}$ of real numbers by, $S_n=\sum _{k=0}^n\frac{1}{\sqrt{n^2+k}} \text{ for }n\ge 1$. Then $\underset{n\to \infty }{\lim }{S}_n$

If $\underset{x\to \frac{\pi }{2}}{\lim }\frac{1-\sin x}{{\left(\frac{\pi }{2}-x\right)}^2}$$=\frac{1}{k}$, then find the value of $k$.

$\underset{x\to \frac{\pi }{4}}{\lim } \frac{1-\tan x}{x-\frac{\pi }{4}}$$=$

If $\underset{x\to a}{\lim } \frac{\cos x-\cos a}{x-a}=k\sin a$, then find the value of $k$.

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

If $\underset{x\to 0}{\lim }\frac{5x+4\sin 3x}{4\sin 2x+7x}=\frac{p}{q}$, the greatest common factor of $p$ and $q$ is $1$, then find the value of $p+q$.

If $\underset{x\to 0}{\lim }\frac{1-\cos 5x}{1-\cos 6x}$$=\frac{k}{36}$, then find the value of $k$.

If $\underset{\mathrm{\theta }\to 0}{\lim }\frac{\sin 4\mathrm{\theta }}{\tan 3\mathrm{\theta }}=\frac{p}{q}$, the greatest common divisor of $p$ and $q$ is $1$, then find the value of $p+q$.

If $\underset{\mathrm{\theta }\to 0}{\lim }\frac{1-\cos 4\mathrm{\theta }}{1-\cos 6\mathrm{\theta }}=\frac{p}{q}$, the greatest common divisor of $p$ and $q$ is $1$, then find the value of $p+q$.

Evaluate the limit: $\underset{x\to 0}{\lim }\left[\frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{x}\right]$

Evaluate the limit $\underset{x\to 0}{\lim }\left[\frac{\sin 2x(\cos 3x-\cos x)}{x^3}\right]$

If $\underset{x\to 0}{\lim }\left[\frac{x^2+1-\cos x}{x\sin x}\right]$$=\frac{k}{2}$, then find the value of $k$.

Evaluate the limit:$\underset{x\to 0}{\lim }\left[\frac{x\tan x}{1-\cos x}\right]$

If $\underset{x\to 0}{\lim }\frac{\sin \left(2+x\right)-\sin \left(2-x\right)}{x}=k\cos 2$, then the value of $k$ is 

Evaluate the limit: $\underset{x\to 0}{\lim }\frac{\cos 3x-\cos 5x}{x^2}$