Limits & Continuity Questions (12)

Let $L_1=\displaystyle\lim_{x\to0}\frac{\tan3x-\tan x}{2\cos4x\sin3x+\sin5x-(2-2\cos4x+\sin3x)}$ and $L_2=\displaystyle\lim_{x\to0^+}\frac{e^{\sec(\tan x^{1012n})}-e}{\sin(x^{2025m})}=\frac{e}{2}$; $n,m\in\mathbb{N}$. Then $\dfrac{1}{4L_1-\frac{m}{n}}$ is

If $x_1=\sqrt{3}$ and $x_{n+1}=\dfrac{x_n}{1+\sqrt{1+x_n^2}}$ for all $n\in\mathbb{N}$, then $\displaystyle\lim_{n\to\infty} 2^n x_n$ is equal to

If $k = \lim_{x\to 1} \sec^{-1}\!\left(\dfrac{\lambda^2}{\ln x} - \dfrac{\lambda^2}{x-1}\right)$ exists, then the minimum value of $[|\lambda|]$ is .......... (where $[\cdot]$ denotes GIF)

Let $f:\mathbb{R}\to\mathbb{R}$ be defined as $f(x)=\displaystyle\lim_{p\to\infty}\frac{\sin\!\left((2x-1)\frac{\pi}{2}\right)\!x^{4p}e^{x^2-1}+x^{4p}}{1+x^{4p+2}-x^{4p}}$. $f(x)$ is continuous for all $x$ in

Let $f:\mathbb{R}\to\mathbb{R}$ be defined as $f(x)=\displaystyle\lim_{p\to\infty}\frac{\sin\!\left((2x-1)\frac{\pi}{2}\right)\!x^{4p}e^{x^2-1}+x^{4p}}{1+x^{4p+2}-x^{4p}}$. $f(x)$ is continuous for all $x$ in

If $x_1=\sqrt{3}$ and $x_{n+1}=\dfrac{x_n}{1+\sqrt{1+x_n^2}}$ for all $n\in\mathbb{N}$, then $\displaystyle\lim_{n\to\infty} 2^n x_n$ is equal to

If $k = \lim_{x\to 1} \sec^{-1}\!\left(\dfrac{\lambda^2}{\ln x} - \dfrac{\lambda^2}{x-1}\right)$ exists, then the minimum value of $[|\lambda|]$ is .......... (where $[\cdot]$ denotes GIF)

If $x_1=\sqrt{3}$ and $x_{n+1}=\dfrac{x_n}{1+\sqrt{1+x_n^2}}$ for all $n\in\mathbb{N}$, then $\displaystyle\lim_{n\to\infty} 2^n x_n$ is equal to

Let $L_1=\displaystyle\lim_{x\to0}\frac{\tan3x-\tan x}{2\cos4x\sin3x+\sin5x-(2-2\cos4x+\sin3x)}$ and $L_2=\displaystyle\lim_{x\to0^+}\frac{e^{\sec(\tan x^{1012n})}-e}{\sin(x^{2025m})}=\frac{e}{2}$; $n,m\in\mathbb{N}$. Then $\dfrac{1}{4L_1-\frac{m}{n}}$ is

If $k = \lim_{x\to 1} \sec^{-1}\!\left(\dfrac{\lambda^2}{\ln x} - \dfrac{\lambda^2}{x-1}\right)$ exists, then the minimum value of $[|\lambda|]$ is .......... (where $[\cdot]$ denotes GIF)

Let $f:\mathbb{R}\to\mathbb{R}$ be defined as $f(x)=\displaystyle\lim_{p\to\infty}\frac{\sin\!\left((2x-1)\frac{\pi}{2}\right)\!x^{4p}e^{x^2-1}+x^{4p}}{1+x^{4p+2}-x^{4p}}$. $f(x)$ is continuous for all $x$ in

Let $L_1=\displaystyle\lim_{x\to0}\frac{\tan3x-\tan x}{2\cos4x\sin3x+\sin5x-(2-2\cos4x+\sin3x)}$ and $L_2=\displaystyle\lim_{x\to0^+}\frac{e^{\sec(\tan x^{1012n})}-e}{\sin(x^{2025m})}=\frac{e}{2}$; $n,m\in\mathbb{N}$. Then $\dfrac{1}{4L_1-\frac{m}{n}}$ is