Limits

If $\lim\limits_{x \to 0}\dfrac{45^x - 9^x - 5^x + 1}{(k^x - 1)(3^x - 1)} = 2$, then the value of $k$ is ...

If $\lim _{x \rightarrow \infty} \frac{(\sqrt{2 x+1}+\sqrt{2 x-1})^8+(\sqrt{2 x+1}-\sqrt{2 x-1})^8\left(P x^4-16\right)}{\left(x+\sqrt{x^2-2}\right)^8+\left(x-\sqrt{x^2-2}\right)^8}$ $=1$, then $P=$

Define $\mathrm{f}(x)=\left\{\begin{array}{cl}\mathrm{b}-\mathrm{a} x & , \text { if } x 2\end{array}\right.$ and if $\lim _{x \rightarrow 2} \mathrm{f}(x)$ exists, then $\frac{\mathrm{a}}{\mathrm{b}}=$

$\lim _{x \rightarrow 2}\left(\frac{5 x-8}{8-3 x}\right)^{\frac{3}{2 x-4}}=$

If $\lim _{\mathrm{x} \rightarrow 5} \frac{\mathrm{x}^{\mathrm{k}}-5^{\mathrm{k}}}{\mathrm{x}-5}=500$, then the value of $\mathrm{k}$, where $\mathrm{k} \in \mathrm{N}$ is

If $f(x)=\frac{\sqrt{\pi}-\sqrt{\operatorname{Cos}^{-1} x}}{\sqrt{1+x}}$ for $x>-1$ then $\lim _{x \rightarrow-1^{+}} f(x)=$

If $\quad \mathbf{f}(x)=\left\{\begin{array}{cl}\frac{1-\cos 4 x}{x^2} & \text { if } x 0\end{array}\right.$ is continuous at $x=0$ Then $a=$

$\lim_{x \to 1} \left[\dfrac{x-2}{x^2-x} - \dfrac{1}{x^3 - 3x^2 + 2x}\right] = $

$\lim _{x \rightarrow 1} \frac{\sqrt{x}-1}{\left(\cos ^{-1} x\right)^2}=$

$$ \lim _{x \rightarrow 0} \frac{\sqrt{1+x \sin x}-\sqrt{\cos x}}{\tan ^2 \frac{x}{2}}= $$

If $\lim\limits_{x \to k} \dfrac{x^3 - k^3}{x^2 - k^2} = \lim\limits_{x \to 0} \dfrac{1 - \cos(2x)}{x \sin x}$, then the value of $k$ is $\ldots$

$\mathrm{S}_1=\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r}, \mathrm{S}_2=\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r}^2$ and $\mathrm{S}_3=\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r}^3$, then the value of $\lim _{n…