Question

*I WILL GIVE A MEDAL* What is the limit as x approaches 0 of tan(x)/2x?

Answer 1

\(\bf lim_{x\to 0}\quad \cfrac{tan(x)}{2x}
\\ \quad \\
\cfrac{tan(x)}{2x}\implies \cfrac{\frac{sin(x)}{cos(x)}}{2x}\implies \cfrac{sin(x)}{cos(x)}\cdot \cfrac{1}{2x}\implies \cfrac{sin(x)}{x}\cdot \cfrac{1}{2cos(x)}\)

any ideas?

Answer 2

First, you need to know these limits all go to 1:

limit as x->0
sin(x)/x

limit as x->0
sin(2x)/2x

limit as x->0
sin(3x)/3x

You may know this theorem.
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Secondly, you need to know your trig identities in order to re-express "tan(x)/x".

Recall tan(x) = sin(x)/cos(x)
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Now we can get started:
lim x->0
tan(x) / x
becomes:
sin(x)/(x * cos(x))
and with a little re-writing:
(sin(x) / x) * (1 / cos(x))

Use the theorem:
(1) * (1 / cos(x))

Plug in x = 0. Cos(0) is 1, therefore:
(1) * (1/1)
= 1

The limit is 1.
You can also check this by graphing "tan(x) / x" on your calculator.