Giải thích các bước giải:
a.Ta có:
$\lim_{x\to3}\dfrac{x-3}{x^2+2x-15}$
$=\lim_{x\to3}\dfrac{x-3}{(x-3)(x+5)}$
$=\lim_{x\to3}\dfrac{1}{x+5}$
$=\dfrac{1}{3+5}$
$=\dfrac18$
b.Ta có:
$\lim_{x\to-\infty}\dfrac{\sqrt{9x^2-3x+1}}{3x-5}$
$=\lim_{x\to-\infty}\dfrac{\sqrt{9x^2-3x+1}:(-x)}{(3x-5):(-x)}$
$=\lim_{x\to-\infty}\dfrac{\sqrt{(9x^2-3x+1):(-x)^2}}{-3+\dfrac5x}$
$=\lim_{x\to-\infty}\dfrac{\sqrt{(9x^2-3x+1):x^2}}{-3+\dfrac5x}$
$=\lim_{x\to-\infty}\dfrac{\sqrt{9-\dfrac3x+\dfrac1{x^2}}}{-3+\dfrac5x}$
$=\dfrac{\sqrt{9-0+0}}{-3+0}$
$=-1$
c.Ta có:
$\lim_{x\to1}\dfrac{\sqrt{x+3}-2}{x-1}$
$=\lim_{x\to1}\dfrac{\dfrac{x+3-2^2}{\sqrt{x+3}+2}}{x-1}$
$=\lim_{x\to1}\dfrac{\dfrac{x-1}{\sqrt{x+3}+2}}{x-1}$
$=\lim_{x\to1}\dfrac{1}{\sqrt{x+3}+2}$
$=\dfrac{1}{\sqrt{1+3}+2}$
$=\dfrac14$
d.Ta có:
$\lim_{x\to2}\dfrac{2-x}{\sqrt{x+7}-3}$
$=\lim_{x\to2}\dfrac{2-x}{\dfrac{x+7-3^2}{\sqrt{x+7}+3}}$
$=\lim_{x\to2}\dfrac{-(x-2)}{\dfrac{(x-2)}{\sqrt{x+7}+3}}$
$=\lim_{x\to2}-(\sqrt{x+7}+3)$
$=-(\sqrt{2+7}+3)$
$=-6$
