trig. limit

**cityismine** · June 9th 2008, 05:34 PM

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.

**galactus** · June 9th 2008, 05:47 PM

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

Divide top and bottom by x:

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

You can see the familiar limits in the num. and den. See what it is now?.

**Mathstud28** · June 9th 2008, 05:52 PM

Originally Posted by cityismine

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.

It is commonly known that $\tan(x)\sim{x}$ and $\sin(x)\sim{x}$

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

Alternatively

you can divide by x to get

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

Evaluating each separately we get

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

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

Thus we have

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

also you could utilize power series but that might be too advanced

**cityismine** · June 9th 2008, 10:15 PM

The division by x method works out perfectly, thanks!

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