Limits of Trig Functions

**mhuggin2** · May 4th 2010, 08:59 PM

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

**pickslides** · May 4th 2010, 09:20 PM

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

**autumn** · May 4th 2010, 10:55 PM

Originally Posted by pickslides

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

YOU mean 1 not 0

**DeMath** · May 4th 2010, 11:33 PM

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

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

**pickslides** · May 5th 2010, 03:36 AM

Originally Posted by autumn

YOU mean 1 not 0

yep, silly typo.

**mhuggin2** · May 5th 2010, 11:58 AM

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

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