Question

Evaluate:

${lim}_{x\to \sqrt{2}}\frac{x^9-{3x}^8+x^6-{9x}^4-{4x}^2-16x+84}{x^5-{3x}^4-4x+12}$

Answer 1

$\underset{x\to \sqrt{12}}{lim}\frac{x^9-3x^8+x^6-9x^4-4x^2-16x+84}{x^5-3x^4-4x+12}$

$\therefore$ Applying L Hospital Rule $\left(\frac{0}{0}from\right)$

$=\underset{x\to \sqrt{2}}{lim}\frac{9x^8-24x^7+6x^5-36x^3-8x-16}{5x^4-12x^3-4}$

$=\underset{x\to \sqrt{2}}{lim}\frac{9(\sqrt{2}{)}^8-24(\sqrt{2}{)}^7+6(\sqrt{2}{)}^5-36(\sqrt{2}{)}^3-8\sqrt{2}-16}{5(\sqrt{2}{)}^4-12(\sqrt{2}{)}^3-4}$

$=\underset{x\to \sqrt{2}}{lim}\frac{9\times 144-192\sqrt{2}+24\sqrt{2}-72\sqrt{2}-8\sqrt{2}-16}{20-24\sqrt{2}-4}$

$=\underset{x\to \sqrt{2}}{lim}\left(\frac{1296-248\sqrt{2}-16}{16-24\sqrt{2}}\right)$

Ans :- $=\frac{1280-248\sqrt{2}}{16-24\sqrt{2}}=\frac{160-31\sqrt{2}}{2-3\sqrt{2}}$