Examine the following function for continuity: $f(x) = \frac{1}{x-5}, x \neq 5$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The given function is $f(x) = \frac{1}{x-5}$,where $x \neq 5$.
For any real number $k$ such that $k \neq 5$,we evaluate the limit:
$\lim_{x \to k} f(x) = \lim_{x \to k} \frac{1}{x-5} = \frac{1}{k-5}$.
Also,the value of the function at $x = k$ is $f(k) = \frac{1}{k-5}$.
Since $\lim_{x \to k} f(x) = f(k)$ for all $k \in \mathbb{R} \setminus \{5\}$,the function $f$ is continuous at every point in its domain.
Therefore,$f$ is a continuous function.

Explore More

Similar Questions

The function $f(x) = [x] \cdot \cos \left( \frac{2x - 1}{2} \right) \pi$,where $[\cdot]$ denotes the greatest integer function,is discontinuous at

If the function $f(x) = \begin{cases} \frac{\tan 4x \times \cos 3x}{x} & , x \neq 0 \\ k & , x = 0 \end{cases}$ is continuous at $x = 0$,then the value of $k$ is equal to . . . . . .

If $f(x) = \begin{cases} \frac{9^x - 2 \cdot 3^x + 1}{\log(1 + 3x) \cdot \tan 2x} & , x \neq 0 \\ a(\log b)^c & , x = 0 \end{cases}$ is continuous at $x = 0$,then $a + b + c =$

If $f(x) = \left(\frac{1+\tan x}{1+\sin x}\right)^{\operatorname{cosec} x}$ is continuous at $x=0$,then $f(0)$ is equal to

If $f(x) = \frac{\log_e(1 + x^2 \tan x)}{\sin x^3}, x \neq 0$ is to be continuous at $x = 0$,then $f(0)$ must be equal to

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