 Continuity
Question

# A function is defined as $\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{n}\to \infty }{\mathrm{lim}} \left\{\begin{array}{l}{\mathrm{cos}}^{2\mathrm{n}}\mathrm{x}, \mathrm{if} \mathrm{x}<0\\ \sqrt[\mathrm{n}]{1+{\mathrm{x}}^{\mathrm{n}}}, \mathrm{if} 0\le \mathrm{x}\le 1\\ \frac{1}{1+{\mathrm{x}}^{\mathrm{n}}}, \mathrm{if} \mathrm{x}>1\end{array}\right\$which of the following does not hold good?

Moderate
Solution

## $\begin{array}{l}\mathrm{f}\left({0}^{-}\right)=\underset{\mathrm{n}\to \infty }{\mathrm{lim}} \left[\underset{\mathrm{n}\to {0}^{-}}{\mathrm{lim}} {\left({\mathrm{cos}}^{2}\mathrm{x}\right)}^{\mathrm{n}}\right]\\ ={\left( \mathrm{a} \mathrm{value} \mathrm{lesser} \mathrm{than} 1\right)}^{\infty }=0\\ \mathrm{f}\left({0}^{+}\right)= \underset{\mathrm{n}\to \infty }{\mathrm{lim}} \left[\underset{\mathrm{n}\to {0}^{+}}{\mathrm{lim}} {\left(1+{\mathrm{x}}^{\mathrm{n}}\right)}^{\frac{1}{\mathrm{n}}}\right]=1\end{array}$Also, f(0) = 1. So, the function is discontinuous at x = 0.$\mathrm{Further}, \mathrm{f}\left({1}^{-}\right)=1; \mathrm{f}\left({1}^{+}\right)=0; \mathrm{f}\left(1\right)=1.$So, the function is discontinuous at x = 1.

Get Instant Solutions  