Question

limit X tends to 3 x square minus 3 upon X square + 3 root 3 x minus 12

Answer 1

Answer:

$\bf\displaystyle\lim_{x\to\sqrt{3}}\:\frac{x^2-3}{x^2+3\sqrt{3}-12}=\frac{2}{5}$

Step-by-step explanation:

Given:

$\displaystyle\lim_{x\to\sqrt{3}}\:\frac{x^2-3}{x^2+3\sqrt{3}-12}$

$=\displaystyle\lim_{x\to\sqrt{3}}\:\frac{x^2-{\sqrt3}^2}{x^2+3\sqrt{3}-12}$

Using $\boxed{a^2-b^2=(a-b)(a+b)}$

$=\displaystyle\lim_{x\to\sqrt{3}}\:\frac{(x-{\sqrt3})(x+{\sqrt3})}{(x+4{\sqrt3})(x-{\sqrt3})}$

$=\displaystyle\lim_{x\to\sqrt{3}}\:\frac{x+{\sqrt3}}{x+4{\sqrt3}}$

$=\frac{{\sqrt3}+{\sqrt3}}{{\sqrt3}+4{\sqrt3}}$

$=\frac{2{\sqrt3}}{5{\sqrt3}}$

$=\frac{2}{5}$

$\implies\boxed{\bf\displaystyle\lim_{x\to\sqrt{3}}\:\frac{x^2-3}{x^2+3\sqrt{3}-12}=\frac{2}{5}}$

Answer 2

Answer:

ur answer is here