Methods of Evaluation of Limits

Let $a_1>a_2>a_3>\dots >a_n>1; p_1>p_2>p_3>\dots >p_n>0;$ such that $p_1+p_2+p_3+\dots +p_n=1$. Also $F\left(x\right)={\left(p_1{a}_1^x+p_2{a}_2^x+\dots +p_n{a}_n^x\right)}^{1/x},$ then $\underset{x\to \infty }{\lim }F\left(x\right)$ equals

If $\underset{x\to 0}{\lim }\frac{1-cosx.cos2x.cos3x\dots \dots \dots \cos nx}{x^2}$ has the value equal to$253$, find the value of $n$ is equal to (where $n\in N$)

The value of $\underset{x\to 0}{\lim }\frac{1+sinx-cosx+\ln \left(1-x\right)}{x·{tan}^{2}x}$ is

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

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

If $S$ be the sum of coefficients in the expansion of ${\left(px+qy-rz\right)}^n$ (where $p, q, r>0$ ), then the value of $\underset{n\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{S}{{\left(S^{\frac{1}{n}}+1\right)}^n},$ is:

Let $f$ and $g$ be two functions on $R$ defined by

$f\left(x\right)=\sqrt{x^2+1}-x$

$g\left(x\right)=\sin \left(\pi e^{1-x}\right)$

Define a function $h:R\to R$ by $h\left(x\right)=\max \left\{f\left(x\right),g\left(x\right)\right\}$. Then what can be said about $\underset{x\to \infty }{\lim }h\left(x\right)?$

$\underset{y\to 0}{\lim }\frac{(y-2)+2\sqrt{1+y+y^2}}{2y}$ is equal to___________

$\underset{x\to 0}{\lim }\frac{\log \left(\sin 7x+\cos 7x\right)}{\sin 3x}$ equals

If $f\left(x\right)=\left|\begin{array}{ccc}\sin x& \cos x& \tan x\\ x^3& x^2& x\\ 2x& 1& 1\end{array}\right|$, then $\underset{x\to 0}{\lim }\frac{f\left(x\right)}{x^2}$ is

Let $g\left(x\right)=x \ln \left(1+\frac{1}{x}\right)$, then$\underset{x\to 0^{+}}{\lim }g\left(x\right)$:

If $f\left(1\right)=2$ and $f^{\prime }\left(1\right)=1$, then value of $\underset{x\to 1}{\lim }\frac{2x-f\left(x\right)}{x-1}$ is

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 $f\left(5\right)=7$ and $f^{\prime }\left(5\right)=7$, then $\underset{x\to 5}{\lim }\frac{xf\left(5\right)-5f\left(x\right)}{x-5}$ is given by

$\underset{x\to \infty }{\lim }\frac{3x^3+2x^2-7x+9}{4x^3+9x-2}$ is equal to

If $f$ is differentiable at $x=1$$,$ then $\underset{x\to 1}{\lim }\frac{x^{2}f\left(1\right)-f\left(x\right)}{x-1}$ is

The value of $\underset{x\to \infty }{\lim }{\left(\frac{x+6}{x+1}\right)}^{x+4}$ is

The value of  $\underset{x\to 0}{\lim }{\left(\frac{1+5x^2}{1+3x^2}\right)}^{\frac{1}{x^2}}$ is

If $f\left(x\right)$ is continuous in $[0, 1]$ and $f\left(\frac{1}{3}\right)=1$ then $\underset{n\to \infty }{\lim }f\left(\frac{n}{\sqrt{9n^2+1}}\right)$

The value of $\underset{x\to 0}{\lim }{\left[\tan \left(x+\frac{\mathrm{\pi }}{4}\right)\right]}^{\frac{1}{x}}$ is