Prove that the function $f(x) = x^{n}$ is continuous at $x = n$,where $n$ is a positive integer.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The given function is $f(x) = x^{n}$.
It is evident that $f$ is defined at all positive integers $n$,and its value at $x = n$ is $f(n) = n^{n}$.
Now,we evaluate the limit of the function as $x$ approaches $n$:
$\lim_{x \to n} f(x) = \lim_{x \to n} (x^{n}) = n^{n}$.
Since $\lim_{x \to n} f(x) = f(n) = n^{n}$,the condition for continuity is satisfied.
Therefore,the function $f(x) = x^{n}$ is continuous at $x = n$,where $n$ is a positive integer.

Explore More

Similar Questions

If the function $f(x) = \begin{cases} \frac{(e^{kx} - 1) \tan kx}{4x^2}, & x \neq 0 \\ 16, & x = 0 \end{cases}$ is continuous at $x = 0$,then $k = . . . . . .$.

If $f: R \rightarrow R$ defined by $f(x) = \begin{cases} \frac{1 + 3 x^2 - \cos 2 x}{x^2}, & x \neq 0 \\ k, & x = 0 \end{cases}$ is continuous at $x = 0$,then $k$ is equal to

If a real-valued function $f(x) = \begin{cases} \frac{2x^2+(k+2)x+9}{3x^2-7x-6} & , \text{for } x \neq 3 \\ l & , \text{for } x=3 \end{cases}$ is continuous at $x=3$ and $l$ is a finite value,then $l-k=$

If $f(x) = \begin{cases} kx + 1, & x \leq \frac{\pi}{2} \\ \sin x, & x > \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$,then $k = $ . . . . . . .

If $f(x) = \begin{cases} -x^2, & \text{when } x \le 0 \\ 5x - 4, & \text{when } 0 < x \le 1 \\ 4x^2 - 3x, & \text{when } 1 < x < 2 \\ 3x + 4, & \text{when } x \ge 2 \end{cases}$,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