Embibe Experts Solutions for Chapter: Limits, Exercise 3: Exercise-3

Let $\alpha ,\beta \in R$ be such that $\underset{x\to 0}{\lim }\frac{{\mathrm{x}}^2\sin \left(\beta x\right)}{\alpha x-\sin x}=1.$ Then $6\left(\alpha +\beta \right)$ equals

Let $f\left(x\right)=\frac{1-x\left(1+|1-x|\right)}{|1-x|}\cos \left(\frac{1}{1-x}\right)$ for $x\ne 1$. Then

$\underset{x\to 0}{\lim }\frac{\left(1-\cos 2x\right)\left(3+\cos x\right)}{x\tan 4x}$ is equal to

Let $p=\underset{x\to 0^{+}}{\lim }{\left(1+{\tan }^2\sqrt{x}\right)}^{\frac{1}{2x}}$,  then ${\log }_e\left(p\right)$ is equal to

$\underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\frac{\cot \left(x\right)-\cos \left(x\right)}{{\left(\mathrm{\pi }-2x\right)}^3}$ equals

For each $t\in R$, let $\left[t\right]$ be the greatest integer less than or equal to $\mathrm{t}.$ Then $\underset{x\to 0^{+}}{\lim }x\left(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+ \dots +\left[\frac{15}{x}\right]\right)$

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

For each $t\in R,$ let $\left[t\right]$ be the greatest integer less than or equal to $t.$ Then,

$\underset{x\to 1^{+}}{lim}\frac{\left(1-\left|x\right|+\sin \left|1-x\right|\right)\sin \left(\frac{\pi }{2}\left[1-x\right]\right)}{\left|1-x\right|\left[1-x\right]}$