List of top Mathematics Questions on limits and derivatives

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

The value of \(\lim_{x \to 0} \frac{6\sin(x) - 2\sin(3x)}{3x^3}\) is equal to 
 

If \(\lim_{x \to 3} \frac{(2x - k)\tan(x - 3)}{x^2 - 6x + 9} = 2\), then the value of 'k' is equal to 
 

\[ \lim_{x\to 0}\frac{\sqrt{1+x}-1}{x}= \]
\[ \lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x= \]
The value of the limit $\lim_{x \to 0} \frac{e^{3x} - e^{-2x}}{\sin 4x}$ is:
The value of the limit \( \lim_{n \to \infty} \left( \frac{1}{1^2 + n^2} + \frac{2}{2^2 + n^2} + \frac{3}{3^2 + n^2} + \cdots + \frac{n}{n^2 + n^2} \right) \) is equal to:
The value of the limit $\lim_{x \to 0} \frac{e^{3x} - e^{-2x}}{\sin 4x}$ is:
The value of $\lim_{x \to \infty} \left(\frac{x+6}{x+1}\right)^{x+4}$ is:
\( \displaystyle \lim_{x \to 2} \left( \frac{1}{x-2} - \frac{2}{x^3-3x^2+2x} \right) = \)
Let the function \[ f(x)=\frac{x-|x|}{x} \] Then
Let the function $f(x)$ be defined as \[ f(x)=\frac{x-|x|}{x} \] then:
If \[ f(x)=\ln\left(\frac{\sin x}{1+\cos x}\right), \] then \(f'(x)\) is equal to:
Given a real valued function \(f\) such that \[ f(x) = \begin{cases} \frac{\tan^2\{x\}}{x^2 - [x]^2} & \text{for } x > 0 \\ 1 & \text{for } x = 0 \\ \sqrt{\{x\} \cot\{x\}} & \text{for } x < 0 \end{cases} \] then
The value of \(\lim_{n \rightarrow \infty} \prod_{r=3}^{n} \frac{r^3 - 8}{r^3 + 8}\) equals to:
$$ \lim_{x \to 0} \frac{1 - \cos x}{x^2} = $$
$$ \lim_{x \to \infty} \left[ \frac{1^x + 2^x + 3^x + \cdots + n^x}{n} \right]^{1/x} = $$
The value of $\lim_{x \to 2} \frac{x^3 + 3x^2 - 9x - 2}{x^3 - x^2 - 4x + 4}$ is ________
The value of $\lim_{x\rightarrow 0}\frac{\sqrt{1+3x}-\sqrt{1-3x}}{x}$ is
Let $y = \dfrac{3x^3 - 2x^2 + x}{|x|}, \; x \ne 0$. Then $\dfrac{dy}{dx}$ at $x=-2$ is equal to:
The value of $\lim_{x \to 1} \dfrac{x - 1}{3\sqrt{x} - 1}$ is equal to:
If $\lim_{x \to 2} \frac{\sin(x^3 - 5x^2 + ax + b)}{(\sqrt{x-1}-1) \log_e(x-1)} = m$ (exists finitely), then find the value of $a + b + m$.
If $\sum_{k=1}^{n} a_k = 6n^3$ then evaluate $\sum_{k=1}^{6} \left( \frac{a_{k+1} - a_k}{36} \right)^2$.
Value of definite integral $I = \int_{0}^{2\sqrt{3}} \log_2(x^2 + 4) dx + \int_{2}^{4} \sqrt{2^x - 4} dx$
The value of $\int_{0}^{3} \frac{e^x + e^{-x}}{[x]!} dx$ is equal to (where $[\, \cdot \,]$ denotes greatest integer function):