Đáp án:
a) $\lim\limits_{x\to 0}=\dfrac{\sqrt{x+9}+\sqrt{x+16}-7}x=\dfrac7{24}$
b) $\lim\limits_{x\to 0}\dfrac{\sqrt{1+x}-1}{\sqrt[3]{1+x}-1}=\dfrac32$
Giải thích các bước giải:
a) $\lim\limits_{x\to 0}=\dfrac{\sqrt{x+9}+\sqrt{x+16}-7}x$
$=\lim\limits_{x\to 0}\dfrac{\sqrt{x+9}-3}x+\lim\limits_{x\to 0}\dfrac{\sqrt{x+16}-4}x$
$=\lim\limits_{x\to 0}\dfrac{x+9-9}{x(\sqrt{x+9}+3)}+\lim\limits_{x\to 0}\dfrac{x+16-16}{x(\sqrt{x+16}+4)}$
$=\lim\limits_{x\to 0}\dfrac{1}{\sqrt{x+9}+3}+\lim\limits_{x\to 0}\dfrac{1}{\sqrt{x+16}+4}$
$=\dfrac1{3+3}+\dfrac1{4+4}$
$=\dfrac16+\dfrac18=\dfrac7{24}$
b) $\lim\limits_{x\to 0}\dfrac{\sqrt{1+x}-1}{\sqrt[3]{1+x}-1}$
$=\lim\limits_{x\to 0}\dfrac{(\sqrt{1+x}-1)(\sqrt{1+x}+1)(\sqrt[3]{(1+x)^2}+\sqrt{1+x}+1))}{(\sqrt{1+x}+1)(\sqrt[3]{1+x}-1)(\sqrt[3]{(1+x)^2}+\sqrt{1+x}+1)}$
$=\lim\limits_{x\to 0}\dfrac{x(\sqrt[3]{(1+x)^2}+\sqrt{1+x}+1)}{x(\sqrt{x+1}+1)}$
$=\lim\limits_{x\to0}\dfrac{\sqrt[3]{(1+x)^2}+\sqrt{1+x}+1}{\sqrt{x+1}+1}$
$\lim\limits_{x\to0}\dfrac{\sqrt[3]{(1+0)^2}+\sqrt{1+0}+1}{\sqrt{0+1}+1}=\dfrac32$
