 Continuity
Question

# , then

Moderate
Solution

## $\begin{array}{l}\mathrm{We} \mathrm{have} \underset{\mathrm{x}\to 0}{\mathrm{lim}} \frac{\mathrm{f}\left(\mathrm{x}\right)-\mathrm{f}\left(0\right)}{\mathrm{x}-0}= \underset{\mathrm{x}\to 0}{\mathrm{lim}} \frac{\mathrm{log} \mathrm{cosx}}{\mathrm{log}\left(1+{\mathrm{x}}^{2}\right)}\\ =\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{\mathrm{log}\left(1-1+\mathrm{cosx}\right)}{\mathrm{log}\left(1+{\mathrm{x}}^{2}\right)}\frac{1-\mathrm{cosx}}{1-\mathrm{cosx}}\\ =\underset{\mathrm{x}\to 0}{\mathrm{lim}}\frac{\mathrm{log}\left\{1-\left(1-\mathrm{cosx}\right)\right\}}{1-\mathrm{cosx}}\frac{1-\mathrm{cosx}}{\mathrm{log}\left(1+{\mathrm{x}}^{2}\right)}\\ =\underset{\mathrm{x}\to 0}{\mathrm{lim}} \frac{\mathrm{log}\left[1-\left(1-\mathrm{cosx}\right)\right]}{1\left(1-\mathrm{cosx}\right)}\frac{2{\mathrm{sin}}^{2}\frac{\mathrm{x}}{2}}{4\left(\frac{\mathrm{x}}{2}\right)}\frac{{\mathrm{x}}^{2}}{\mathrm{log}\left(1+{\mathrm{x}}^{2}\right)}=-\frac{1}{2}\\ \mathrm{Hence}, \mathrm{f}\left(\mathrm{x}\right) \mathrm{is} \mathrm{differetiable} \mathrm{x}=0.\end{array}$Hence, (2') and (4) are the correct answers.

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