1. $\lim_{x \to 0^+} \sqrt{\dfrac{1}{x}+2} - \sqrt{\dfrac{1}{x}}$
2. $\lim_{x \to 0^+} \dfrac{\sqrt{2-x^2}}{x}$
3. $\lim_{x \to 0} x \sin (\dfrac{1}{x})$
Don't know how to solve. Can someone help?
For the first use $\displaystyle (a- b)\frac{a+ b}{a+ b}= \frac{a^2- b^2}{a+ b}$. Then multiply both numerator and denominator by $\displaystyle \sqrt{x}$.
For the second, write it as $\displaystyle \sqrt{\frac{2- x^2}{x^2}}$
For the third, $\displaystyle -1\le sin(1/x)\le 1$.
First:
$\lim_{x \to 0^+} \dfrac{\dfrac{1}{x}+2-\dfrac{1}{x}}{\sqrt{\dfrac{1}{x}+2}+\sqrt{\dfrac{1 }{x}}}
= \lim_{x \to 0^+} \dfrac{2\sqrt{x}}{\sqrt{1+2x}+1}
= \lim_{x \to 0^+}\dfrac{0}{2}=0
$
Second:
$\lim_{x \to 0^+}\sqrt{\dfrac{2- x^2}{x^2}}$
Conclude that the limit does not exist, because I can't manipulate the expression any more?
Third:
Still confused.