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 $\underset{x\to 0}{\lim }\frac{x^{6000}-{\left(\sin x\right)}^{6000}}{x^2{\left(\sin x\right)}^{6000}}$ is

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

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

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

Let n be an odd integer, if  $\sin n\theta ={\displaystyle \sum _{r=0}^n{b}_r{\sin }^r\theta }$ , for every value of  $\theta$ , then

If $\underset{x\to 0}{\lim }\frac{e^{ax}-\cos \left(bx\right)-\frac{cx{e}^{-cx}}{2}}{1-\cos \left(2x\right)}=17$, then $5a^2+b^2$ is equal to

Let $a_1,a_2,a_3,....,a_n$ be $\mathrm{n}$ positive consecutive terms of an arithmetic progression. If $d>0$ is its common difference, then $\underset{n\to \infty }{\lim }\sqrt{\frac{d}{n}}\left(\frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\dots +\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}\right)$ is

$\underset{n\to \infty }{\lim }\left\{\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)\left(2^{\frac{1}{2}}-2^{\frac{1}{5}}\right)....\left(2^{\frac{1}{2}}-2^{\frac{1}{2n+1}}\right)\right\}$ is equal to

$\underset{x\to -8}{\lim }\left(\frac{\sqrt[3]{x}+2}{x+8}\right)=$

The value of $\underset{x\to 0}{\lim }\left[\frac{\left(e^x-x-1\right)\left(x-\sin x\right)\ln \left(1+x\right)}{x^6}\right]$ is equal to

Find the value of $\underset{x\to 0}{\lim }{\left(\cos x\right)}^{co{t}^{2}x}$.

$\underset{x\to 1}{\lim }\left(1-x\right)\tan \left(\frac{\mathrm{\pi x}}{2}\right)$ is equal to

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

$\underset{x\to -2}{\lim }\left(\frac{x^9+512}{x^5+32}\right)=$

The value of the limit $\underset{x\to 0^{+}}{\lim }{\left(\sin x\right)}^{\sqrt{x}}{\left(e^x+x\right)}^{\frac{1}{x}}$ is

Evaluate the limit: $\underset{x\to 0^{+}}{\lim }{\left(\sin x\right)}^{\sqrt{x}}{\left(e^x+x\right)}^{\frac{1}{x}}$.

Find the value of the limit $\underset{x\to 0}{\lim }{\left[{\sin }^2\left(\frac{\mathrm{\pi }}{2-ax}\right)\right]}^{{\sec }^2\left(\frac{\mathrm{\pi }}{2-\mathrm{ax}}\right)}$

Evaluate $\underset{x\to 1}{\lim }\left(\frac{10}{1-x^{10}}-\frac{3}{1-x^3}\right)$