Hi guys!

Someone in the tutoring center at my school asked me to do a limit problem for their calc 1 class, and I came up with the following solution. I would like someone to come up with an alternate solution, I don't really like mine, since I applied a theorem I believe to be true but don't recall ever seeing it done. I ended up with the right answer though, but I don't know if that was a coincidence.

Here's the "theorem" I used:

Theorem:If $\displaystyle \lim_{x \to a} f(x) = L$ then $\displaystyle \lim_{x \to a} \frac {1}{f(x)} = \frac {1}{L}$. Provided $\displaystyle L < \infty$ and $\displaystyle L \neq 0$

Proof:Assume $\displaystyle \lim_{x \to a } f(x) = L$.

Then $\displaystyle \lim_{x \to a} \frac {1}{f(x)} = \frac {\lim_{x \to a} 1}{\lim_{x \to a} f(x)}$ ..........since we are not dealing with infinite limits, the limit of the quotient is equal to the quotient of the limits (that's true right? )

Thus we have $\displaystyle \lim_{x \to a} \frac {1}{f(x)} = \frac {1}{L}$

QED

Now here's the question, and my solution.

Find $\displaystyle \lim_{ \theta \to 0} \frac {\sin \theta}{ \theta + \tan \theta}$

(Yeah, I know, you guys thought I'd have a much harder problem for you, well no, i'm only smart enough to tackle easy problems)

Solution:

$\displaystyle \lim_{ \theta \to 0} \frac {\sin \theta}{ \theta + \tan \theta} = \lim_{ \theta \to 0} \frac { \frac {1}{ \csc \theta}}{ \theta + \frac { \sec \theta }{ \csc \theta}}$

....................$\displaystyle = \lim_{ \theta \to 0} \frac { \frac {1}{ \csc \theta}}{ \frac { \theta \csc \theta + \sec \theta}{ \csc \theta}}$

....................$\displaystyle = \lim_{ \theta \to 0} \frac {1}{ \theta \csc \theta + \sec \theta}$

....................$\displaystyle = \lim_{ \theta \to 0} \frac {1}{ \frac { \theta}{ \sin \theta} + \sec \theta}$

....................$\displaystyle = \frac {\lim_{\theta \to 0} 1}{\lim_{\theta \to 0} \frac { \theta }{ \sin \theta} + \lim_{\theta \to 0} \sec \theta}$

Now recall that $\displaystyle \lim_{\theta \to 0} \frac { \sin \theta }{ \theta } = 1$.

Applying the theorem I attempted to prove above we have that $\displaystyle \lim_{ \theta \to 0} \frac { \theta }{ \sin \theta} = 1$ as well (This is the part I'm not sure about, I've never this limit defined as that before)

So, $\displaystyle \frac { \lim_{ \theta \to 0} 1}{ \lim_{ \theta \to 0} \frac { \theta }{ \sin \theta} + \lim_{ \theta \to 0} \sec \theta} = \frac {1}{1 + 1}$

Therefore $\displaystyle \boxed { \lim_{ \theta \to 0 } \frac {\sin \theta}{ \theta + \tan \theta} = \frac {1}{2}}$

Can anyone come up with a better and/or shorter method? (L'Hopital's is not allowed).

Thanks