Which of the following functions is not continuous at $x = 0$?
- A$f(x) = \begin{cases} (1 + 2x)^{1/x}, & x \neq 0 \\ e^2, & x = 0 \end{cases}$
- B$f(x) = \begin{cases} \sin x - \cos x, & x \neq 0 \\ -1, & x = 0 \end{cases}$
- C$f(x) = \begin{cases} \frac{e^{1/x} - 1}{e^{1/x} + 1}, & x \neq 0 \\ -1, & x = 0 \end{cases}$
- D$f(x) = \begin{cases} \frac{e^{5x} - e^{2x}}{\sin 3x}, & x \neq 0 \\ 1, & x = 0 \end{cases}$
Explore More
Similar Questions
If $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} \frac{2 \sin x-\sin 2 x}{2 x \cos x}, & \text{if } x \neq 0 \\ a, & \text{if } x=0 \end{cases}$, then the value of $a$ so that $f$ is continuous at $x=0$ is
If $f(x) = \begin{cases} \frac{x - 1}{2}, & 0 \leqslant x < 1 \\ 1/2, & 1 \leqslant x < 2 \end{cases}$ and $g(x) = (2x + 1)(x - k) + 3$ for $0 \leqslant x < \infty$,then $g(f(x))$ will be continuous at $x = 1$ if $k$ is equal to:
If $f(x) = \left(\frac{1+x}{1-x}\right)^{\frac{1}{x}}$ is continuous at $x = 0$,then $f(0) = $
Let $f: R \rightarrow R$ be defined by $f(x)=\begin{cases} \alpha+\frac{\sin [x]}{x}, & x>0 \\ 2, & x=0 \\ \beta+\left[\frac{\sin x-x}{x^3}\right], & x < 0 \end{cases}$. If $f$ is continuous at $x=0$,find the value of $\alpha + \beta$.
DifficultTS EAMCET 2012
View SolutionLet $f(x) = [x^2 - x] + |-x + [x]|$,where $x \in R$ and $[t]$ denotes the greatest integer less than or equal to $t$. Then,$f$ is
DifficultJEE MAIN 2023
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo