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function with denominator zero

 
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Jul27-09, 10:08 PM   #1
 

function with denominator zero


1. The problem statement, all variables and given/known data

This is a limit problem but what I need to figure out is simpler so I thought I'd post it under pre-calc. The question is:

Find the limit:

lim as x approaches 0 of (tan 2x)/x

2. Relevant equations



3. The attempt at a solution

Since x is in the denominator I know that I must re-write (tan 2x)/x so that the denominator doesn't equal zero. I also know that 0 is a root of both the numerator and denominator, but I don't know how to re-write such an equation. Any help? Thanks!
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Jul27-09, 10:16 PM   #2
 
Use L'Hopital's Rule
Jul27-09, 10:32 PM   #3

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If you haven't done l'Hopital yet, do you know lim x->0 tan(x)/x=1 or lim x->0 sin(x)/x=1? Then you could just do the variable substitution u=2x.
Jul28-09, 05:45 PM   #4
 

function with denominator zero


Why can't we obtain the limit here by squeezing?

I'll show you a similar example with tan(x) instead of (tan 2x):

[tex]\lim_{x \to 0} \frac{tanx}{x} = \lim_{x \to 0} (\frac{sin x}{x}.\frac{1}{cos x})[/tex]

[tex](\lim_{x \to 0} \frac{sin x}{x})(\lim_{x \to 0} \frac{1}{cos x}) = (1)(1) =1[/tex]

In your question you must do some manipulation for the 2 in tan2x.
Jul28-09, 06:28 PM   #5
 
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Quote by roam View Post
Why can't we obtain the limit here by squeezing?

I'll show you a similar example with tan(x) instead of (tan 2x):

[tex]\lim_{x \to 0} \frac{tanx}{x} = \lim_{x \to 0} (\frac{sin x}{x}.\frac{1}{cos x})[/tex]

[tex](\lim_{x \to 0} \frac{sin x}{x})(\lim_{x \to 0} \frac{1}{cos x}) = (1)(1) =1[/tex]

In your question you must do some manipulation for the 2 in tan2x.
You're not actually doing any "squeezing" here--just using the fact that [tex]\lim_{x \to 0} \frac{sin x}{x} = 1[/tex]
This limit is often proved by the "squeeze-play" theorem, but can be done other ways.
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