Function with denominator zero

[3.141592654]

Homework Statement

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

Homework Equations

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!

 

[Feldoh]

Use L'Hopital's Rule

 

[Dick]

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.

 

[roam]

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

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

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

(\lim_{x \to 0} \frac{sin x}{x})(\lim_{x \to 0} \frac{1}{cos x}) = (1)(1) =1

In your question you must do some manipulation for the 2 in tan2x.

 

[Mark44]

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

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

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

(\lim_{x \to 0} \frac{sin x}{x})(\lim_{x \to 0} \frac{1}{cos x}) = (1)(1) =1

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 \lim_{x \to 0} \frac{sin x}{x} = 1
This limit is often proved by the "squeeze-play" theorem, but can be done other ways.