Methods of Evaluation of Limits

$\underset{x\to 0}{\lim }\frac{8}{x^8}\left[1-\cos \frac{x^2}{2}-\cos \frac{x^2}{4}+\cos \frac{x^2}{2}\cos \frac{x^2}{4}\right]$

What is $\underset{x\to 1}{\lim }\frac{x^9-1}{x^3-1}$ equal to?

$\underset{y\to 0}{\lim }\frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4}$

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

If $f\left(x\right)=\frac{x-3}{x+4}, g\left(x\right)=-\frac{2\left(2x+1\right)}{x^2+x-12}$ and $r\left(x\right)=\frac{2}{x-3}$, then the value of $\underset{x\to 3}{\lim }\left\{f\left(x\right)+g\left(x\right)+r\left(x\right)\right\}$ is

$\underset{x\to 0}{\lim }\left(\frac{{\sin }^{2}x}{\sqrt{2}-\sqrt{1+\cos x}}\right)$ is equal to

$\underset{n\to \infty }{\lim }\prod _{k=0}^n\left(1+\frac{2}{{45}^{2^k}+{45}^{-2^k}}\right)=$

$\underset{x\to -\infty }{\lim }\left(\sqrt{x^2-2x-1}-\sqrt{x^2-7x+3}\right)$ is equal to

$\underset{x\to 1}{\lim }\left[\frac{\left(x-1\right)\left(x^2-1\right)\left(x^3-1\right)\left(x^4-1\right)}{{\left(\left(x-1\right)\left(x^2-1\right)\right)}^2}\right]$ is equal to

$\underset{x\to \infty }{\lim }{x}^3\left[\sqrt{x^2+\sqrt{x^4+1}}-\sqrt{2}x\right]=$

$\frac{d}{dx}\left(\underset{y\to 2}{\lim }\frac{1}{y-2}\left(\frac{1}{x}-\frac{1}{x+y-2}\right)\right)=$

$\underset{t\to 0}{\lim }\left(\frac{\left(2t-3\right)\left(t-2\right)}{t}-\frac{3\left(t+2\right)}{t}\right)$ is equal to

If the function $f\left(x\right)=\frac{\left(1-\cos 5x\right)\sin 5x}{x^2\sin 3x},x\ne 0$ is continuous at $x=0$, then $f\left(0\right)$ is _____ .

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

$\underset{y\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{\sqrt{3+y^3}-\sqrt{3}}{y^3}=$

Evaluate the limit $\underset{x\to 0}{\lim }\frac{\sqrt{1+x}-1}{x}$

Evaluate $\underset{x\to \sqrt{2}}{\lim }\left(\frac{x^4-4}{x^2+3\sqrt{2}x-8}\right)$

$\underset{x\to 0}{\lim }\frac{{\left(x+2\right)}^{{\displaystyle \frac{1}{3}}}-{\left(2\right)}^{\frac{1}{3}}}{x}=$

Evaluate the limit $\underset{x\to 1}{\lim }\left(\frac{x-\sqrt[n+1]{{\displaystyle \frac{\left(n+1\right)x^n-1}{n}}}}{{\left(x-1\right)}^2}\right)$

$\underset{n\to \infty }{\lim }\sum _{k=1}^n\frac{k^3+6k^2+11k+5}{(k+3)!}=$