Find lim θ → 0: sinθ / θ + tanθ

[TsAmE]

Homework Statement

Find lim θ → 0: sinθ / θ + tanθ

Homework Equations

None.

The Attempt at a Solution

lim sinθ / θ + tanθ
θ → 0

= lim sinθ / θ + (sinθ / cosθ)
θ → 0

= lim sinθ / cosθ + sinθ / cosθ
θ → 0

= lim sinθcosθ / cosθ + sinθ
θ → 0

=sin(0)cos(0) / cos(0) + sin(0)

= 0 * 1 / 1 + 0

= 0

The right answer is a 1/2. I don't know what I could have done wrong, as my method produces a reasonable answer of 0

 

[ystael]

If you can't use TeX, please use parentheses to make things unambiguous. Are you trying to compute \lim_{\theta\to 0} \frac{\sin\theta}{\theta} + \tan\theta or \lim_{\theta\to 0} \frac{\sin\theta}{\theta + \tan\theta}?

 

[physicsman2]

It should be the second one, ystael, because the limit of the first is 1.

 

[ystael]

That's clear to me; I'm trying to get the OP to state his/her problem correctly.

 

[TsAmE]

Its the second one. Sorry but I don't know how to use TeX. How can I use it so that I can not be ambiguous?

 

[ystael]

Notation: If you need to write it on one line, you can write \lim_{\theta\to 0} (\sin\theta)/(\theta + \tan\theta); the top or bottom of a fraction must be parenthesized when it contains an operator that has a space in it (usually, anything except multiplications). Even \sin\theta / (\theta + \tan\theta) (which is what I wrote in the first draft of this response) leaves some ambiguity: this could be taken to mean \sin \frac{\theta}{\theta + \tan\theta}.

Your mistake above was in the transition from the second to the third line. In fact, because of the ambiguity in your fractions, I'm not sure what the third line is even supposed to mean, but that plain \theta in the denominator just disappeared.

Hint: Sometimes when you have a problem with an inconvenient fraction, you can make your life easier by computing with its reciprocal instead. Some argument may be necessary afterward to prove that taking 1 over your whole reasoning is justified.