Embibe Experts Solutions for Chapter: Limits, Exercise 1: JEE Main - 12th April 2019 Shift 2

If $\underset{x\to 0}{\lim }\frac{\alpha e^x+\beta e^{-x}+\gamma \sin x}{x{\sin }^{2}x}=\frac{2}{3}$, where $\alpha ,\beta ,\gamma \in R$, then which of the following is NOT correct?

$\underset{t\to 0}{\lim }{\left(1^{\frac{1}{{\sin }^{2}t}}+2^{\frac{1}{{\sin }^{2}t}}+3^{\frac{1}{{\sin }^{2}t}}......n^{\frac{1}{{\sin }^{2}t}}\right)}^{{\sin }^{2}t}$ is equal to

The set of values of $a$ for which $\underset{x\to a}{\lim }\left(\left[x-5\right]-\left[2x+2\right]\right)=0$, where, $\left[\zeta \right]$ denotes the greatest integer less than or equal to $\zeta$ is equal to

The value of

$\underset{n\to \infty }{\lim }\frac{1+2-3+4+5-6+\dots +(3n-2)+(3n-1)-3n}{\sqrt{2n^4+4n+3-}\sqrt{n^4+5n+4}}$ is

Let $x=2$ be a root of the equation $x^2+px+q=0$ and $f\left(x\right)=\left\{\begin{array}{cc}\frac{1-\cos \left(x^2-4px+q^2+8q+16\right)}{{\left(x-2p\right)}^4},& x\ne 2p\\ 0,& x=2p\end{array}\right.$. Then $\underset{x\to 2p^{+}}{\lim }\left[f\left(x\right)\right]$
where $\left[·\right]$ denotes greatest integer function, is

$\underset{x\to 0}{\lim }\frac{48}{x^4}{\int }_0^x\frac{t^3}{t^6+1}dt$  is equal to 

Let $f,g$ and $h$ be the real valued functions defined on $\mathrm{ℝ}$ as $f\left(x\right)=\left\{\begin{array}{cc}\frac{x}{\left|x\right|},& x\ne 0\\ 1,& x=0\end{array}\right., g\left(x\right)=\left\{\begin{array}{cc}\frac{\sin \left(x+1\right)}{\left(x+1\right)},& x\ne -1\\ 1,& x=-1\end{array}\right.$and $h\left(x\right)=2\left[x\right]-f\left(x\right)$, where $\left[x\right]$ is the greatest integer $\le x$. Then the value of $\underset{x\to 1}{\lim }g\left(h\left(x-1\right)\right)$ is

$\underset{x\to \infty }{\lim }\frac{{\left(\sqrt{3x+1}+\sqrt{3x-1}\right)}^6+{\left(\sqrt{3x+1}-\sqrt{3x-1}\right)}^6}{{\left(x+\sqrt{x^2-1}\right)}^6+{\left(x-\sqrt{x^2-1}\right)}^6}{x}^3$