Đáp án:
$a)$ $\lim_{x\to 4}\dfrac{x^2+3x-4}{x^2+4x}=\dfrac 34$
$b)$ $\lim_{x\to +\infty}\dfrac{15}{x^2-6}=0$
$c)$ $\lim_{x\to 1}\dfrac{3-\sqrt{2x+7}}{x^2-1}=-\dfrac 16$
$a)$
$\lim_{x\to 4}\dfrac{x^2+3x-4}{x^2+4x}$
$=\lim_{x\to 4}\dfrac{(x+4)(x-1)}{x(x+4)}$
$=\lim_{x\to 4}\dfrac{x-1}{x}$
$=\dfrac{4-1}4$
$=\dfrac 34$
$b)$
$\lim_{x\to +\infty}\dfrac{15}{x^2-6}$
$=0$
$c)$
$\lim_{x\to 1}\dfrac{3-\sqrt{2x+7}}{x^2-1}$
$=\lim_{x\to 1}\dfrac{9-2x-7}{(x^2-1)(3+\sqrt{2x+7})}$
$=\lim_{x\to 1}\dfrac{-2(x-1)}{(x+1)(x-1)(3+\sqrt{2x+7})}$
$=\lim_{x\to 1}\dfrac{-2}{(x+1)(3+\sqrt{2x+7})}$
$=\dfrac{-2}{2.6}$
$=-\dfrac 16$
