Question

(i) Evaluate ${lim}_{x\to \frac{\pi }{6}}\frac{\sqrt{3}sinx-cosx}{x-\frac{\pi }{6}}$.

(ii) If $f\left(x\right)=1+x+\frac{x^2}{2}+...+\frac{x^{100}}{100},$ then find the value of f'(1).

Answer 1

(i) Let $E={lim}_{x\to \frac{\pi }{6}}\frac{\sqrt{3}sinx-cosx}{x-\frac{\pi }{6}}={lim}_{x\to \frac{\pi }{6}}\frac{2(\frac{\sqrt{3}}{2}sinx-\frac{1}{2}cosx)}{x-\frac{\pi }{6}}$

$={lim}_{x\to \frac{\pi }{6}}\frac{2(sinxcos\frac{\pi }{6}-sin\frac{\pi }{6}cosx)}{x-\frac{\pi }{6}}$

$={lim}_{x\to \frac{\pi }{6}}\frac{2\left[sin(x-\frac{\pi }{6})\right]}{x-\frac{\pi }{6}}$

On putting $x-\frac{\pi }{6}=h$ as $x\to \frac{\pi }{6}$, then $h\to 0$

$\therefore E={lim}_{h\to 0}\frac{2sinh}{h}=2\times 1=2$

(ii) Given, $f\left(x\right)=1+x+\frac{x^2}{2}+...+\frac{x^{100}}{100}$

On differentiating both sides w.r.t. x, we get

$f^{\prime }\left(x\right)=0+1+\frac{2x}{2}+...+\frac{100x^{99}}{100}$

$\Rightarrow f^{\prime }\left(x\right)=1+x+x^2+...+x^{99}$

$\Rightarrow f^{\prime }\left(1\right)=1+1+1^2+...+1^{99}$

$=100$