Question

Evaluate $\underset{x\to \sqrt{10}}{lim}\frac{\sqrt{7-2x}-(\sqrt{5}-\sqrt{2})}{x^2-10}$

Or

Differentiate $x^{2}cosx$ by first principle.

Answer 1

We have,$\underset{x\to \sqrt{10}}{lim}\frac{\sqrt{7-2x}-(\sqrt{5}-\sqrt{2})}{x^2-10}$

$=\underset{x\to \sqrt{10}}{lim}\frac{\sqrt{7-2x}-(\sqrt{5}-\sqrt{2})}{x^2-10}$

$=\underset{x\to \sqrt{10}}{lim}\frac{\sqrt{7-2x}-\sqrt{7-2\sqrt{10}}}{x^2-10}$

$=\underset{x\to \sqrt{10}}{lim}\frac{\sqrt{7-2x}-\sqrt{7-2\sqrt{10}}}{x^2-10}\times =\underset{x\to \sqrt{10}}{lim}\frac{\sqrt{7-2x}+\sqrt{7-2\sqrt{10}}}{\sqrt{7-2x}+\sqrt{7-2\sqrt{10}}}$

$=\underset{x\to \sqrt{10}}{lim}\frac{(7-2x)-(7-2\sqrt{10})}{(x+\sqrt{10})(x-\sqrt{10})\left(\sqrt{7-x}\right)+\sqrt{7-2\sqrt{10}}}$

$=\underset{x\to \sqrt{10}}{lim}\frac{-2x+2\sqrt{10}}{(x+\sqrt{10})(x-\sqrt{10})\left(\sqrt{7-x}\right)+\sqrt{7-2\sqrt{10}}}$

$=\underset{x\to \sqrt{10}}{lim}\frac{-2(x-\sqrt{10}}{(x+\sqrt{10})(x-\sqrt{10})\left(\sqrt{7-x}\right)+\sqrt{7-2\sqrt{10}}}$

$\underset{x\to \sqrt{10}}{lim}\frac{-2}{(x+\sqrt{10})\left(\sqrt{7-2x}\right)+\sqrt{7-x\sqrt{10}}}$

$\underset{x\to \sqrt{10}}{lim}\frac{-2}{2\sqrt{10}\left(2\sqrt{7-2\sqrt{10}}\right)}=\frac{-1}{2\sqrt{10}\sqrt{(\sqrt{5}-\sqrt{2}{)}^2}}$

$=\frac{-1}{2\sqrt{10(\sqrt{5}-\sqrt{2})}}=\frac{-1}{2\sqrt{10}(\sqrt{5}-\sqrt{2})}\times \frac{(\sqrt{5}+\sqrt{2})}{(\sqrt{5}+\sqrt{2})}$

$=\frac{-(\sqrt{5}+\sqrt{2})}{2\sqrt{10}\times 3}=-\frac{(\sqrt{5}+\sqrt{2})}{6\sqrt{10}}$

Or

$Letf\left(x\right)=x^{2}cosx.thenf(x+h)=(x+h{)}^{2}cos(x+h)$

$\therefore f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$

$=\underset{h\to 0}{lim}\frac{(x^2+2hx+h^2)cos(x+h)-x^{2}cosx}{h}$

$=\underset{h\to 0}{lim}\frac{x^2\left[cos\right(x+h)-cosx]+h(2x+h)\left(cos\right(x+h\left)\right)}{h}$

$=\underset{h\to 0}{lim}\frac{x^2(-2)sin\left(\frac{x+h+x}{2}\right)sin\left(\frac{x+h-x}{2}\right)}{h}+2xcosx$

$=\underset{h\to 0}{lim}\frac{-2x^{2}sin(x+\frac{h}{2})sin\frac{h}{2}}{2\times \frac{h}{2}}+2xcosx$

$=-x^{2}sinx+2xcosx[\because =\underset{h\to 0}{lim}\frac{sinh}{h}=1]$