Limits Questions - JEE

The value of $\lim_{x \to 0} \frac{e^{(1+x)^{1/x}} - e}{\tan x}$ is

Let \[ f(x) = \begin{cases} xe^{-\left(\frac{1}{|x|}+\frac{1}{x}\right)}, & x \neq 0 \\ 0, & x = 0 \end{cases} \] Then $f(x)$ is

If A = limx→0 sin−1(sin x) cos−1(cos x) and B = limx→0 [|x|] x , then:

\(f(x)=\begin{cases}\max\{|x|,x^2\} & |x|\le 2\\ 8-2|x| & 2. S = non-diff in (-4,4).

Evaluate $\lim_{x \to 0} \dfrac{64^x - 32^x - 16^x + 4^x + 2^x - 1}{\left[\sqrt{(15 + \cos x)} - 4\right]\sin x}$

$f(x+y)=f(x)+f(y)+xy^2+x^2y$, $\lim_{x\to 0}f(x)/x=1$. Find $f'(3)$.

$\displaystyle\lim_{x\to\frac{\pi}{2}}\left(\dfrac{1}{\left(x-\frac{\pi}{2}\right)}\int_{x}^{\frac{\pi}{2})^3}\cos\!\left(\dfrac{1}{t^3}\right)dt\right)$ is equal to

If $ f'(x) + (f'(x))^2 + (f'(x))^3 + (f'(x))^4 + \cdots \infty = e^x $, where $ f'(x) \in (-1, 1) $ and $ f(0) = 0 $, then the value of $ \lim_{x \to 0} (1 + f(x))^{\frac{1}{x}} $ is equal to:

⎧ ⎪ 3x, x < 0 Let f (x) = ⎨ min{1 + x + [x], x + 2[x]}, 0 \le x \le 2 ⎩ ⎪ 5, x > 2, where [.] denotes greatest integer function. If \alpha and \beta are the number of points, where f is not continuous and is not differentiable, respectively, then \alpha + \beta equals __________

Let $f(x)$ be a real differentiable function such that $f(0) = 1$ and $f(x+y) = f(x)f'(y) + f'(x)f(y)$ for all $x, y \in \mathbb{R}$. Then $\displaystyle\sum_{n=1}^{100} \log_e f(n)$ is equal to:

Let $a$ be the sum of all coefficients in the expansion of $(1-2x+2x^2)^{2023}(3-4x^2+2x^3)^{2024}$ and $b=\displaystyle\lim_{x\to0}\left(\dfrac{\int_0^x\dfrac{\log(1+t)}{t^{2024}+1}\,dt}{x^2}\right)$. If the equations $cx^2+dx+e=0$ and $2bx^2+ax+4=0$ have a common root, where $c,d,e\in\mathbb{R}$, then $d:c:e$ equals

Let $f:\mathbb{R}\to\mathbb{R}$ be a function given by $f(x)=\begin{cases}\dfrac{1-\cos2x}{x^2}, & x<0\\ \alpha, & x=0\\ \dfrac{\beta\sqrt{1-\cos x}}{x}, & x>0\end{cases}$ where $\alpha,\beta\in\mathbb{R}$. If $f$ is continuous at $x=0$, then $\alpha^2+\beta^2$ is equal to:

Non-differentiable points of $f(x)=|2x+1|-3|x+2|+|x^2+x-2|$:

If $f(x) = g(x)|(x-1)(x-2)\cdots(x-10)| - 2$ is derivable for all $x \in R$, where $g(x) = ax^9 + bx^6 + cx^3 + d$, $a,b,c,d \in R$, then $f'(-1)$ is equal to:

Find a for which limx→0+ (1−x)1/x−e−1 xa is non-zero.

If $y = \ln\\!\left\{\dfrac{x + \sqrt{a^2 + x^2}}{a}\right\}$, then the value of $\dfrac{dy}{dx}$ is:

Consider $f(x) = x \ln x$ and $g(x) = e^{2x}$. Let $a$ and $b$ be two values of $x$ satisfying $f(x) = g(x)$ with $a If \(h(x) = \frac{f(x)}{g(x)}$, then $h'(a)$ equals:

If the function $f(x)=\dfrac{\sin3x+\alpha\sin x-\beta\cos3x}{x^3}$, $x\in\mathbb{R}$, is continuous at $x=0$, then $f(0)$ is equal to:

$f(x)=|x^2-2x-3|\cdot e^{|9x^2-12x+4|}$ non-diff at:

Let $f: [1, 10] \to \mathbb{Q}$ be a continuous function and $f(1) = 10$, then $f(10)$ is equal to

Consider the function $$f(x)=\begin{cases}\dfrac{a(7x-12-x^2)}{b|x^2-7x+12|} & ,\; x<3\\2^{\sin\left(\frac{x-3}{x-[x]}\right)} & ,\; x>3\\b & ,\; x=3\end{cases}$$ where $[x]$ denotes the greatest integer $\le x$. If $S$ denotes the set of all ordered pairs $(a,b)$ such that $f(x)$ is continuous at $x=3$, then the number of elements in $S$ is:

Let $f(x) = \lim_{n \to \infty} \displaystyle\sum_{r=0}^{n}\left(\frac{\tan(x/2^{r+1}) + \tan^3(x/2^{r+1})}{1 - \tan^2(x/2^{r+1})}\right)$. Then $\lim_{x \to 0} \dfrac{e^x - e^{f(x)}}{x - f(x)}$ is equal to:

Multiplicative functional equation, $f'(1)=2024$. ODE:

Find $\lambda$ for $f''(0)$ to exist.

g(x)=|f(x)|+f(|x|). In (-2,2), g is:

$\lim_{x \to \infty} \dfrac{(2x^2 - 3x + 5)(3x-1)^{x/2}}{(3x^2 + 5x + 4)\sqrt{(3x+2)^x}}$ is equal to:

$f$ triangle function, $g=f(x+2)-f(x-2)$. $n+m$?

If the function $f(x)=\begin{cases}\dfrac{1}{|x|} & ,\; |x|\ge 2\\ax^2+2b & ,\; |x|<2\end{cases}$ is differentiable on $\mathbb{R}$, then $48(a+b)$ is equal to ______.

The value of $\displaystyle\lim_{x\to\frac{\pi}{2}}\frac{4(x-\pi)\cos^2 x}{\pi(\pi-2x)\tan\!\left(x-\dfrac{\pi}{2}\right)}$ is equal to:

Let $f(x) = \begin{cases} \left[1 + \ln(c^2 + c + 1)\tan^2(x-1)\right]^{\frac{1}{(\ln x)^2}}, & x \neq 1 \\ 3c, & x = 1 \end{cases}$ where $c \in R$. If $\lim_{x \to 1} f(x)$ exists but $f(x)$ is discontinuous at $x = 1$, then $c$ can take the value:

201. Let $f: \mathbb{R} \to \mathbb{R}$ be a function such that for all $x, y \in \mathbb{R}$, $|f(x) - f(y)| \leq 6|x - y|^2$. If $f(3) = 6$, then $f(6)$ is equal to:

Consider the function $f:(0,2)\to\mathbb{R}$ defined by $f(x)=\frac{x}{2}+\frac{2}{x}$ and the function $g(x)$ defined by $$g(x)=\begin{cases}\min\{f(t)\},\; 0<t\le x & ,\; 0<x\le 1\\\frac{3}{2}+x & ,\; 1<x<2\end{cases}$$ Then

The set of points where $ f(x) = \dfrac{x}{1+|x|} $ is differentiable, is

Let $f: R \to R$ defined by $f(x) = x^3 + 3x + 1$ and $g$ be the inverse of $f$, then the value of $g''(5)$ equals:

Same $f$ and $f_1(x)=|f(|x|)|$. Number of points in $[-2,10]$ where $f_1$ is NOT differentiable:

$f=\max\{1+x+[x],2+x,x+2[x]\}$ on $[0,2]$. $(m+n)^2+2=$

f piecewise. m+n = non-cont + non-diff in (-2,2):

200. If $y = 2\tan^{-1}\!\left(\dfrac{\sqrt{1+x^2}-1}{x}\right)$, then the value of $\dfrac{d^2y}{dx^2}$ at $x = 2$ is:

Let $f(x) = x|x|$ and $g(x) = \sin x$. Then $g \circ f$ is:

Let $\displaystyle\lim_{n\to\infty}\left(\frac{n}{\sqrt{n^4+1}}-\frac{2n}{(n^2+1)\sqrt{n^4+1}}+\frac{n}{\sqrt{n^4+16}}-\frac{8n}{(n^2+4)\sqrt{n^4+16}}+\cdots+\frac{n}{\sqrt{n^4+n^4}}-\frac{2n\cdot n^2}{(n^2+n^2)\sqrt{n^4+n^4}}\right)$ be $\dfrac{\pi}{k}$, using only the principal values of the inverse trigonometric functions. Then $k^2$ is equal to

If $x^n = x^2 + x + 1$, then $\lim_{n \to \infty} x^n$ (as $n \to \infty$, $x \to 1$) equals:

Let a be a positive real number. Let f : R →R and g : (a, ∞) →R be defined by f(x) = sin πx 12 , g(x) = 2 loge(√x −√a) loge(e √x −e √a) . Then the value of limx→a+ f(g(x)) is

If $y^2 = P(x)$ which is a polynomial of degree 3, then $2\dfrac{d}{dx}\left(y^3 \dfrac{d^2y}{dx^2}\right)$ equals

If limx→0 eax−cos(bx)−cxe−cx 2 1−cos 2x = 17, find 5a2 + b2.

$\lim_{x \to 0} \frac{\sqrt{\frac{1}{2}(1-\cos 2x)}}{x}$ is equal to

Let f(x) = ( sin x x ∈Z 0 x /∈Z and g(x) =      x2 + 1 x ̸= 0, 2 4 x = 0 5 x = 2 , then:

limx→1− √π− √ 2 sin−1 x √1−x is equal to:

Given $(2x)^{2y} = 4e^{2x-2y}$, then $\dfrac{dy}{dx}(1 + \log_e 2x)^2$ equals:

Limit check for piecewise f(x):

Let $ S = \{t \in R : f(x) = |x - \pi| \cdot (e^{|x|} - 1)\sin|x| $ is not differentiable at $t\}$. Then the set S is equal to:

Limits Questions (1092)