lim (x -> 0) sin(x)/(x+tan(x))

Can someone show me step by step method of solving this. I tried it, but my answer doesn't match the one in the book.

Printable View

- Jun 9th 2008, 05:34 PMcityisminetrig. limit
lim (x -> 0) sin(x)/(x+tan(x))

Can someone show me step by step method of solving this. I tried it, but my answer doesn't match the one in the book. - Jun 9th 2008, 05:47 PMgalactus
$\displaystyle \lim_{x\to\rightarrow{0}}\frac{sin(x)}{x+tan(x)}$

Divide top and bottom by x:

$\displaystyle \lim_{x\to\rightarrow{0}}\frac{\frac{sin(x)}{x}}{1 +\frac{tan(x)}{x}}$

You can see the familiar limits in the num. and den. See what it is now?. - Jun 9th 2008, 05:52 PMMathstud28
It is commonly known that $\displaystyle \tan(x)\sim{x}$ and $\displaystyle \sin(x)\sim{x}$

$\displaystyle \therefore\lim_{x\to{0}}\frac{\sin(x)}{x+\tan(x)}= \lim_{x\to{0}}\frac{x}{2x}=\frac{1}{2}$

Alternatively

you can divide by x to get

$\displaystyle \lim_{x\to{0}}\frac{\frac{\sin(x)}{x}}{1+\frac{\ta n(x)}{x}}$

Evaluating each separately we get

$\displaystyle \lim_{x\to{0}}\frac{\sin(x)}{x}=\lim_{x\to{0}}\fra c{\sin(x)-\sin(0)}{x-0}=\bigg(\sin(x)\bigg)'\bigg|_{x=0}=1$

and $\displaystyle \lim_{x\to{0}}\frac{\tan(x)}{x}=\lim_{x\to{0}}\fra c{\tan(x)-\tan(0)}{x-0}=\bigg(\tan(x)\bigg)'\bigg|_{x=0}=1$

Thus we have

$\displaystyle \frac{1}{1+1}=\frac{1}{2}$

also you could utilize power series but that might be too advanced - Jun 9th 2008, 10:15 PMcityismine
The division by x method works out perfectly, thanks!