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  4. If f(x)=(√2+ cos x-1/(π-x)2) is continuous at x=π then f(π)=....

Q. If $f(x)=\frac{\sqrt{2+\cos x}-1}{(\pi-x)^{2}}$ is continuous at $x=\pi$ then $f(\pi)=$....

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Solution:

since $f$ is continuous at $x=\pi, f(\pi)=\underset{x \rightarrow \pi}{\text{lt}}\left[\frac{\sqrt{2+\cos x}-1}{(\pi-x)^{2}}\right]$
$=\underset{x \rightarrow \pi}{\text{lt}}\left[\frac{1(-\sin x)}{2 \sqrt{2+\cos x}} / 2(\pi-x)(-1)\right](\text { by L.H. Rule })$
$=\underset{x \rightarrow \pi}{\text{lt}}\left[\frac{\sin (x)}{4 \sqrt{2+\cos x}}\left(\frac{1}{\pi-x}\right)\right]=\frac{1}{4} \cdot \frac{1}{\sqrt{2-1}} \cdot \underset{x \rightarrow \pi}{\text{lt}}\left(\frac{\sin (\pi-x)}{(\pi-x)}\right)(\text { by L.H. Rule })$
$=\frac{1}{4} \cdot(1)=\frac{1}{4}$

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