Limits Questions (1092)

The limit limx→∞ (x+1)10+(x+2)10+···+(x+100)10 x10+1010 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

Let f(x) = sin{x} x2+ax+b. If f(5+) and f(3+) exist finitely and are non-zero, find (a + b):

Let $k\in\mathbb{Z}$. If $\displaystyle\lim_{x\to 0^+}(\sin(kx)+\cos x+x)^{2/x}=e^6$, then the value of $k$ is:

Let $f(x)=\begin{pmatrix}x^{\ln(2x-1)}\cdot\dfrac{(x)^{x\ln x}}{(x^{e^x-2}-1)\cdot x^e\cdot\sin x}\end{pmatrix}$. Then the right-hand limit of $f(x)$ at $x=0$ equals:

Let $k\in\mathbb{Z}$. If $\displaystyle\lim_{x\to 0^+}(\sin(kx)+\cos x+x)^{2/x}=e^6$, then the value of $k$ is:

Let \(f(x)\) be differentiable function on the interval \((0, \infty)\) such that \(f(1) = 1\) and \(\lim_{t \to x} \frac{t^3 f(x) - x^3 f(t)}{t^2 - x^2} = \frac{1}{2}\) for all \(x > 0\), then \(f(x)\) is:

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

The value of limx→0  tan π 4 + x
1/x is:

\(f(x)=\begin{cases}a+\sin(\sin x) & x\ge 0\\ \ln(\cos x)+bx & x differentiable at \(x=0\). Then:

If $f(x) = \begin{cases} x+1; & x > 1 \\ 0; & x = 1 \text{ then } f'(0) \text{ equals to} \\ -3x; & x < 1 \end{cases}$

Let $[x]$ denote the greatest integer function, and let $m$ and $n$ respectively be the numbers of points, where the function $f(x) = [x] + |x-2|$, $-2 < x < 3$, is not continuous and not differentiable. Then $m + n$ is equal to:

The value of $\lim_{x \to 0} \frac{\sin(2x + \tan 2x)}{x^3}$ is equal to

Let $[x]$ denote the greatest integer function and $f(x)=\max\{1+x+[x],\ 2+x,\ x+2[x]\}$, $0\leq x\leq 2$, where $m$ is the number of points where $f$ is not continuous and $n$ be the number of points in $(0,2)$ where $f$ is not differentiable. Then $(m+n)^2+2$ is equal to

Let x = 2 be a root of the equation x^2 + px + q = 0 and f(x) = \begin{cases} \dfrac{1-\cos(x^2-4px+q^2+8q+16)}{(x-2p)^4}, & x \neq 2p \\ 0, & x = 2p \end{cases}. Then \lim_{x \to 2p^+} [f(x)], where [.] denotes greatest integer function, is

If $\lim_{x \to \infty} \left(1 + p^2 + q^2(x)\right) = e^p$, then

If $f(x) = (x-1)(x-2)(x-3)(x-4)(x-5)$, then the value of $f'(5)$ is equal to

If limx→0[1 + x ln(1 + b2)]1/x = 2b sin2 θ, b > 0 and θ ∈(−π, π], then the value of θ is:

If limx→1 x4−1 x−1 = limx→k x3−k3 x2−k2 , then k is:

Same \(f(x)=\text{tgn}(x)\cdot\sin(x)\cdot|x|\). Points of non-differentiability in \([-10,10]\):

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 constants \(a\), \(b\) and \(c\) so that \(\lim_{x\to 0}\dfrac{axe^x - b\log(1+x) + cxe^{-x}}{x^2\sin x} = 2\) then find \((a+b-c)/8\).

$\displaystyle\lim_{x\to0}\left(\dfrac{1-\cos^2(3x)}{\cos^3(4x)}\right)\left(\dfrac{\sin^3(4x)}{(\ln(2x+1))^5}\right)$ is equal to

\(f(x)=\begin{cases}e^{x^2}+x^3+1 & x>0\\ ax^2+bx+2 & x\le 0\end{cases}\) is differentiable at \(x=0\). Then:

The value of limx→0 ln(sin 3x) ln(sin x) is:

The value of $\lim_{x \to 0} \frac{1-\cos(\tan x)}{x^2 \sin(2\sin(x)\cdot \cos(\frac{1}{x}))}$ is equal to

\(f(x)=\begin{cases}\sin^{-1}x+a\cos\pi x & -1 is differentiable in \((-1,1)\). Find \(2a+b\).

Define $f: [0, \pi] \to \mathbb{R}$ by $f(x) = \begin{cases} \tan^2 x + \sqrt{2\sin^2 x + 3\sin x + 4 - \sqrt{\sin^2 x + 6\sin x + 2}} & x \neq \pi/2 \\ k & x = \pi/2 \end{cases}$ is continuous at $x = \pi/2$, then $k$ is equal to:

If \[\lim_{x \to 0} \frac{x(1 + a\cos x) - b\sin x}{x^3} = 1\] then

If $x^2 + y^2$ and $z = x + 3t, y = 2x - t$, then $\frac{dz}{dy}$ is equal to (where $t$ is a constant)

Let f, g and h be the real valued functions defined on \mathbb{R} as f(x) = \begin{cases} \frac{x}{|x|}, & x \neq 0 \\ 1, & x = 0 \end{cases}, g(x) = \begin{cases} \frac{\sin(x+1)}{(x+1)}, & x \neq -1 \\ 1, & x = -1 \end{cases} and h(x) = 2[x] - f(x), where [x] is the greatest integer \leq x. Then the value of \lim_{x \to 1} g(h(x-1)) is

If $f(x) = \begin{cases} \frac{\sqrt{a+x}-\sqrt{a-x}}{x} & -1 \leq x < 0 \\ \frac{3x+2}{x-3} & 0 \leq x \leq 1 \end{cases}$ is continuous in $[-1, 1]$, then the value of $a$ is

If \(f(x)\) is odd linear polynomial with \(f(1) = 1\), then \[\lim_{x \to 0} \frac{2^{f(\tan x)} - 2^{f(\sin x)}}{x^2 f(\sin x)}\] is

Let $f(x)=\begin{cases}x^2\sin\left(\dfrac{1}{x}\right), & x\neq0\\0, & x=0\end{cases}$. Then at $x=0$:

Let f(x) = ( 1 + 2x a 0 ≤x

If $\displaystyle\lim_{x\to0}\frac{e^{(a-1)x}+2\cos bx+(c-2)e^{-x}}{x\cos z-\log_e(1+z)}=2$, then $a^2+b^2+c^2$ is equal to:

Which of the following limits equal 1/2?

Given, $\lim_{x \to \infty} \frac{2x + 3^{x+1}}{(2x+3)^2 + (2x+4)^3}$

If $y = \log_a 2 + \log_c 10 + \log_c 10 + \log_{10} 10$, then $\frac{dy}{dx}$ is equal to

If limx→a 2x− √ x2+3a2 √x+a− √ 2a = √ 2 (where a ∈R+), then a is equal to:

$\lim_{x \to \infty} \sqrt[3]{(x+a)(x+b)(x+c)} - x =$

$\lim_{x \to 0} \frac{1 + \sin x - \cos x - \sin(x) \cdot x}{x \sin^2 x}$

If limx→λ 2 −λ x
λ tan(πx/2λ) = 1 e, then λ is equal to:

The function $f(x) = \frac{1 - \sin x + x}{1 - \sin x \cos x}$ is not defined at $x = \pi$. The value of $f(\pi)$, so that $f(x)$ is continuous at $x = \pi$, is

If \(\lim_{x \to 0} \left(1 + ax + bx^2\right)^{2/x} = e^3\), then

\(f(x)=\max\{x,x^2\}\). Non-diff set:

If f(x) is a polynomial of least degree such that limx→0  1 + f(x)+x2 x2 1/x = e2, then f(2) is:

The function $f(x) = [x] + \sqrt{\{x\}}$, where $[.]$ denotes the greatest integer function and $\{.\}$ denotes the fractional part function respectively, is discontinuous at

Matrix Match:(P) f diff at x=3, f'(3)=2: \(\lim_{h\to 0}\frac{f(3+h^2)-f(3-h^2)}{2h^2}\)(Q) f(-x)=f(x), f'(0) exists: f'(0)(R) \(f(x)=\frac{x}{1+e^{1/x}}\) (x≠0), 0 (x=0): Lf'(0)(S) f=max{a-x, a+x, b}, 0<a<b: non-diff points

S = non-diff of \(|2-|x-3||\). Find \(\sum_{x\in S}f(f(x))\).