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This problem is driving me crazy
- Thread starter dsdlewis4
- Start date
Hello, dsdlewis4!
You're expected to know that: \(\displaystyle \:\lim_{x\to0}\left(\frac{\sin x}{x}\right)\:=\:1\)
We have: \(\displaystyle \:\lim_{x\to0}\left(\frac{x}{\sin x}\,+\,\frac{\tan x}{\sin x}\right)\;=\;\lim_{x\to0}\left(\frac{x}{\sin x}\,+\,\frac{1}{\cos x}\right)\)
\(\displaystyle \;\;\;=\;\lim_{x\to0}\left(\frac{x}{\sin x}\right)\:+\:\lim_{x\to0}\left(\frac{1}{\cos x}\right)\)
\(\displaystyle \;\;\;=\;1\,+\,\frac{1}{1}\;=\;2\)
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: \(\displaystyle \:\lim_{x\to0}\left(\frac{x}{\sin x}\,+\,\frac{\tan x}{\sin x}\right)\;=\;\lim_{x\to0}\left(\frac{x}{\sin x}\,+\,\frac{1}{\cos x}\right)\)
\(\displaystyle \;\;\;=\;\lim_{x\to0}\left(\frac{x}{\sin x}\right)\:+\:\lim_{x\to0}\left(\frac{1}{\cos x}\right)\)
\(\displaystyle \;\;\;=\;1\,+\,\frac{1}{1}\;=\;2\)
Daniel_Feldman
Full Member
- Joined
- Sep 30, 2005
- Messages
- 252
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!
(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!