evaluating limits of trig functions # 3

**skeske1234** · August 19th 2009, 12:37 PM

Hi. I have a few questions I am stuck on.

Evaluate each of the following limits.

1. lim (1-cosx)/ tanx
x-> 0

2. lim sinx/ (x+sinx)
x-> 0

And for the last one, #3, I am wondering If I have the right answer and the back is wrong. Take a look please.
3. lim (sin2x)/(2x^2+x)
x->0
My answer: 1
back of the book: 2

Please let me know. Thank you for your help!

**red_dog** · August 19th 2009, 01:12 PM

1) $\lim_{x\to 0}\frac{1-\cos x}{\tan x}=\lim_{x\to 0}\frac{1-\cos x}{\sin x}\cdot\lim_{x\to 0}\cos x=$

$=\lim_{x\to 0}\frac{2\sin^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\f rac{x}{2}}=\lim_{x\to 0}\tan\frac{x}{2}=0$

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

3) $\lim_{x\to 0}\frac{\sin 2x}{2x^2+x}=\lim_{x\to 0}\frac{\sin 2x}{x}\cdot\frac{1}{2x+1}=2\cdot 1=2$

**skeske1234** · August 19th 2009, 01:50 PM

Originally Posted by red_dog

1) $\lim_{x\to 0}\frac{1-\cos x}{\tan x}=\lim_{x\to 0}\frac{1-\cos x}{\sin x}\cdot\lim_{x\to 0}\cos x=$

$=\lim_{x\to 0}\frac{2\sin^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\f rac{x}{2}}=\lim_{x\to 0}\tan\frac{x}{2}=0$

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

3) $\lim_{x\to 0}\frac{\sin 2x}{2x^2+x}=\lim_{x\to 0}\frac{\sin 2x}{x}\cdot\frac{1}{2x+1}=2\cdot 1=2$

ok, i get all of them except for number 1.

after you divide the limit into two parts, the step after that with the [2sin^(2) x/2]/ [2sin(x/2)cos(x/2)] <<<<< i dont get this step. What rule or formula did you use here?

**Chris L T521** · August 19th 2009, 01:54 PM

Originally Posted by skeske1234

ok, i get all of them except for number 1.

after you divide the limit into two parts, the step after that with the [2sin^(2) x/2]/ [2sin(x/2)cos(x/2)] <<<<< i dont get this step. What rule or formula did you use here?

He's using the following identities:

$\sin^2\left(\tfrac{\alpha}{2}\right)=\frac{1-\cos \alpha}{2}$

$\sin\left(2\alpha\right)=2\sin\alpha\cos\alpha$

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