In order that the function $f(x) = (x + 1)^{1/x}$ is continuous at $x = 0$,$f(0)$ must be defined as

  • A
    $f(0) = 0$
  • B
    $f(0) = e$
  • C
    $f(0) = 1/e$
  • D
    $f(0) = 1$

Explore More

Similar Questions

Let $[x]$ denote the greatest integer function and $f(x) = \max\{1+x+[x], 2+x, x+2[x]\}$ for $0 \leq x \leq 2$. Let $m$ be the number of points in $[0, 2]$ 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:

Show that the function defined by $f(x)=\cos \left(x^{2}\right)$ is a continuous function.

Let $f(x) = \begin{cases} (x - 1)^{\frac{1}{2 - x}}, & x > 1, x \neq 2 \\ k, & x = 2 \end{cases}$. The value of $k$ for which $f$ is continuous at $x = 2$ is

If $f(x) = \begin{cases} \frac{(e^{ax}-1) \log(1+x)}{\sin^2 x} & \text{if } x > 0 \\ 2 & \text{if } x = 0 \\ \frac{\cos 4x - \cos bx}{\tan^2 x} & \text{if } x < 0 \end{cases}$ is continuous at $x = 0$,then $\sqrt{b^2 - a^2} = $

If $f(x) = \begin{cases} (1 + 2x)^{1/x}, & x \ne 0 \\ e^2, & x = 0 \end{cases}$,then:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo