Question

(i) Evaluate $\underset{x\to 1}{lim}\frac{(2x-3)(\sqrt{x}-1)}{2x^2+x-3}$ (ii) Differentiate $\frac{x+sinx}{x+cosx}$ with respect to x.

Answer 1

$=\underset{x\to 1}{lim}\frac{(2x-3)(\sqrt{x}-1)}{2x^2+x-3}$

$=\underset{x\to 1}{lim}\frac{(2x-3)(\sqrt{x}-1)(\sqrt{x}+1)}{(2x^2+x-3)(\sqrt{x}+1)}$

$\left[multiplyingnumeratoranddenominatorby\right(\sqrt{x}+1\left)\right]$

$=\underset{x\to 1}{lim}\frac{(2x-3)(x-1)}{(2x^2+x-3)(\sqrt{x}+1)}$

$=\underset{x\to 1}{lim}\frac{(2x-3)(x-1)}{(2x+3)(x-1)(\sqrt{x}+1)}$

$=\underset{x\to 1}{lim}\frac{2x-3}{(2x+3)(\sqrt{x}+1)}$

$=\frac{2\times 1-3}{(2\times 1+3)(\sqrt{1}+1)}=-\frac{1}{10}$

$\left(ii\right)Wehave,\frac{d}{dx}\left(\frac{x+sinx}{x+cosx}\right)$

$=\frac{(x+cosx)\frac{d}{dx}(x+sinx)-(x+sinx)\frac{d}{dx}(x+cosx)}{(x+cosx{)}^2}$

$=\frac{(x+cosx)(1+cosx)-(x+sinx)(1-sinx)}{(x+cosx{)}^2}$

$=\frac{(x+xcosx+cosx+co{s}^{2}x-x-sinx+xsinx+si{n}^{2}x)}{(x+cosx{)}^2}$

$=\frac{x(cosx+sinx)=cosx-sinx+(co{s}^{2}x+si{n}^{2}x)}{(x+cosx{)}^2}$

$=\frac{x(cosx+sinx)+cosx-sinx+1}{(x+cosx{)}^2}$