# Thread: Limits of Trig Functions

1. ## Limits of Trig Functions

I've been working on this problem for a long time now and I haven't been able how to figure it out. It is:

lim sin x/x+tan x *as x approaches 0*
The answer is 1/2 but i dont understand how to get there

2. Consider $\displaystyle \lim_{x\to 0 }\frac{\sin x }{x} = 0$

3. Originally Posted by pickslides
Consider $\displaystyle \lim_{x\to 0 }\frac{\sin x }{x} = 0$
YOU mean 1 not 0

4. Originally Posted by mhuggin2
I've been working on this problem for a long time now and I haven't been able how to figure it out. It is:

lim sin x/x+tan x *as x approaches 0*
The answer is 1/2 but i dont understand how to get there
Divide the numerator and denominator by x

$\displaystyle \lim_{x\to0}\frac{\sin{x}}{x+\tan{x}}=\lim_{x\to0} \frac{\dfrac{\sin{x}}{x}}{1+\dfrac{1}{\cos{x}}\cdo t\dfrac{\sin{x}}{x}}=\frac{1}{1+\dfrac{1}{1}\cdot1 }=\frac{1}{2}.$

5. Originally Posted by autumn
YOU mean 1 not 0
yep, silly typo.

6. ## Thanks

OMG thank you SO much. It all makes so much sense now