lim(x->0) (x+sin(x))/x

[RunningMan]

Hello,
It's been a long time since I've taken a Calculus course, and I'm just starting to review the material. I've done about 25 limit questions so far without any difficulty, but this one has me stumped. I'm pretty sure the answer is 1 (from x/x), but I'm not sure how to show that lim (x->0) sin(x)/x = 0.
Any suggestions?

 

[pka]

Recall that \(\displaystyle \lim _{x \to 0} \frac{{\sin (x)}}{x} = 1\).
The answer is 2.

 

[RunningMan]

Thanks for your reply. How do I show it though? Or, is it just something that is generally "known"?

 

[galactus]

Rewrite as \(\displaystyle \frac{x}{x}+\frac{sinx}{x}\)

Now, we can easily see that x/x is 1 and the famous limit sinx/x=1, so it's 2.

If you want to prove the famous limit \(\displaystyle \lim_{x\to{0}}\frac{sinx}{x}=1\), that probably isn't necessary.

But, the Squeezing Theorem is generally used to prove that.

 

[RunningMan]

Thank you to both of you!

 

[Dr. Flim-Flam]

The limit as x approaches 0 for (x+sin(x))/x gives the indeterminant form 0/0.

Hence L'Hopital to the rescue. d[x+sin(x)]/dy = 1+cos(x) and d[x]/dy = 1.

Ergo the limit as x approaches 0 = (1+1)/1 = 2.