Find the limit of [sin(x)]/[x + tan(x)] as x -> 0

[MarkSA]

I'm having some trouble with this one:

Find the limit as x approaches 0 of:
[sin(x)]/[x + tan(x)]

First thing I did was to change tan(x) to sin(x)/cos(x). I then used the LCD of the denominator and ended up with:

[sin(x)cos(x)]/[(x * cos(x)) + sin(x)]

I'm not sure where to go from here. I can't see any further simplification using trig identities or otherwise. Can anyone help?

 

[Deleted member 4993]

Do you know L'Hospital's rule?

 

[soroban]

Re: Find the limit..

Hello, MarkSA!

We're expected to know: \(\displaystyle \L\:\lim_{x\to0}\frac{\sin x}{x} \:=\:1\;\) and \(\displaystyle \L\:\lim_{x\to0}\,\frac{x}{\sin x} \:=\:1\)

\(\displaystyle \L\lim_{x\to0}\,\frac{\sin x}{x\,+\,\tan x }\)

Divide top and bottom by \(\displaystyle \sin x:\)

. . \(\displaystyle \L\lim_{x\to0}\,\frac{1} {\frac{x}{\sin x}\,+\,\sec x} \;=\;\frac{1}{1\,+\,1} \;=\;\frac{1}{2}\)

 

[MarkSA]

Thanks for the help!

I knew that lim sinx/x = 1 and lim x/sinx = 1, but I couldn't get the equation into the form to make use of that. I'm so used to multiplying the top and bottom of a fraction by something that it rarely crosses my mind to divide the top and bottom by something. I need to get into the habit of recognizing that.

We haven't yet learned L'Hospitals rule.