Question

(i) Evaluate ${lim}_{x\to 0}\frac{{10}^x-2^x-5^x+1}{xtanx}$.
(ii) Differentiate $\frac{1+tanx}{1-tanx}$ with respect to x.

Answer 1

(i) We have ${lim}_{x\to 0}\frac{{10}^x-2^x-5^x+1}{xtanx}$

$={lim}_{x\to 0}\frac{5^x{.2}^x-2^x-5^x+1}{xtanx}={lim}_{x\to 0}\frac{2^x(5^x-1)-1(5^x-1)}{xtanx}$

$={lim}_{x\to 0}\frac{(2^x-1)(5^x-1)}{xtanx}$

$={lim}_{x\to 0}\frac{2^x-1}{x}\times {lim}_{x\to 0}\frac{5^x-1}{x}\times {lim}_{x\to 0}\frac{x}{tanx}$

$=log2\times log5\times 1$

$=\left(log2\right)\left(log5\right)$

(ii) Let $y=\frac{1+tanx}{1-tanx}$

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

$\frac{dy}{dx}=\frac{(1-tanx)\frac{d}{dx}(1+tanx)-(1+tanx)\frac{d}{dx}(1-tanx)}{(1-tanx{)}^2}$

$=\frac{(1-tanx)(se{c}^{2}x-(1+tanx\left)\right(-se{c}^{2}x\left)\right)}{(1-tanx{)}^2}$

$=\frac{se{c}^{2}x(1-tanx+1-tanx)}{(1-tanx{)}^2}=\frac{2se{c}^{2}x}{(1-tanx{)}^2}$