This problem is driving me crazy

[dsdlewis4]

Lim (x + tan x) / (sin x)
x>0

any suggestions[/u]

 

[soroban]

Hello, dsdlewis4!

You're expected to know that: $\displaystyle \:\lim_{x\to0}\left(\frac{\sin x}{x}\right)\:=\:1$

$\displaystyle \lim_{x\to0}\left(\frac{x\,+\,\tan x}{\sin x}\right)$

We have: $\displaystyle \:\lim_{x\to0}\left(\frac{x}{\sin x}\,+\,\frac{\tan x}{\sin x}\right)\;=\;\lim_{x\to0}\left(\frac{x}{\sin x}\,+\,\frac{1}{\cos x}\right)$

$\displaystyle \;\;\;=\;\lim_{x\to0}\left(\frac{x}{\sin x}\right)\:+\:\lim_{x\to0}\left(\frac{1}{\cos x}\right)$

$\displaystyle \;\;\;=\;1\,+\,\frac{1}{1}\;=\;2$

 

[Gene]

Do you know L'Hopital's rule?

 

[Daniel_Feldman]

Let's break it up.

(x+tanx)/sinx=(x/sinx)+(tanx/sinx)


From the squeeze theorem:

Lim (sinx/x)=1
x--->0

So Lim (sinx/x)=1/1=1 (note the reciprocals x/sinx and sinx/x).
x--->0


Now, tanx/sinx=(sinx/cosx)/sinx=1/cosx

Lim (1/cosx)=1/cos(0)=1
x--->0

So

Lim (x + tan x) / (sin x) =1+1=2
x>0



Edit:Soroban beats me!