$\lim_{x \to 0,y\to 0} \frac{\ (x\tan y+4)^\frac12-2}{y \sin x}=?$

Question

I have a question:

$$\lim_{x \to 0,y\to 0} \frac{\ (x\tan y+4)^\frac12-2}{y \sin x}=?$$

Thanks ahead:)

Use $(1+x)^n-1\sim nx$ when $x\rightarrow0$ – Lion Jun 14 '14 at 13:22 — Lion, Jun 14 '14 at 13:22
Is there a simple method to solve this question? – Paul Jun 14 '14 at 13:31 — Paul, Jun 14 '14 at 13:31

score 3 · Accepted Answer · answered Jun 14 '14 at 13:26

We have

$$(x\tan y+4)^{\frac12}=2\left(1+\frac{x\tan y}{4}\right)^{\frac12}\sim2\left(1+\frac{x\tan y}{8}\right)$$ so $$ \frac{\ (x\tan y+4)^\frac12-2}{y \sin x}\sim\frac14\frac{x}{\sin x}\frac{\tan y}{y}\xrightarrow{(x,y)\to(0,0)}\frac14$$

$\lim_{x \to 0,y\to 0} \frac{\ (x\tan y+4)^\frac12-2}{y \sin x}=?$

1 Answers1