# Finding the limit

• Jun 27th 2013, 09:45 PM
math604
Finding the limit
Hi, I need to find the limit of this trigonometric function.

Attachment 28684

I used the trig identity for the cosine, the one for cos(a+b), and i get -sinh.

Attachment 28685

Now I'm stuck. I think I should use the squeeze theorem for this, but I'm not exactly sure. Even if I do use it, the h is still on the bottom and it makes the equation undefined.

Any help would be much appreciated.
• Jun 27th 2013, 10:36 PM
Prove It
Re: Finding the limit
Draw a unit circle with the vertical tangent, showing clearly the lengths \displaystyle \displaystyle \begin{align*} \sin{(\theta)}, \cos{(\theta)} \end{align*} and \displaystyle \displaystyle \begin{align*} \tan{(\theta)} \end{align*} for some angle \displaystyle \displaystyle \begin{align*} \theta \end{align*} in the first quadrant. Can you see that the area of the circular sector is a little greater than the area of the smaller triangle, and a little less than the area of the larger triangle. But as \displaystyle \displaystyle \begin{align*} \theta \to 0 \end{align*}, the areas will all also \displaystyle \displaystyle \begin{align*} \to 0 \end{align*}. That means

\displaystyle \displaystyle \begin{align*} \frac{1}{2}\sin{(\theta)}\cos{(\theta)} \leq \frac{\theta}{2\pi} \, \pi \, r^2 &\leq \frac{1}{2}\tan{(\theta)} \\ \frac{1}{2}\sin{(\theta)}\cos{(\theta)} \leq \frac{\theta}{2} &\leq \frac{\sin{(\theta)}}{2\cos{(\theta)}} \textrm{ since } r = 1 \textrm{ in a unit circle and } \tan{(\theta)} = \frac{\sin{(\theta)}}{\cos{(\theta)}} \\ \sin{(\theta)}\cos{(\theta)} \leq \theta &\leq \frac{\sin{(\theta)}}{\cos{(\theta)}} \\ \cos{(\theta)} \leq \frac{\theta}{\sin{(\theta)}} &\leq \frac{1}{\cos{(\theta)}} \textrm{ (we don't need to worry about the direction the inequality signs take because }\sin{(\theta)} > 0 \textrm{ in the first quadrant)}\\ \frac{ 1 }{ \cos{ ( \theta) } } \geq \frac{ \sin{ ( \theta ) } }{ \theta } &\geq \cos{ ( \theta )} \\ \cos{(\theta)} \leq \frac{ \sin{ ( \theta ) } }{ \cos{ (\theta) } } &\leq \frac{1}{ \cos{ ( \theta ) } } \end{align*}

Now as \displaystyle \displaystyle \begin{align*} \theta \to 0, \, \cos{(\theta)} \to 1 \end{align*} and \displaystyle \displaystyle \begin{align*} \frac{1}{\cos{(\theta)}} \to 1 \end{align*}. Since \displaystyle \displaystyle \begin{align*} \frac{\sin{(\theta)}}{\theta} \end{align*} is sandwiched between two quantities that go to 1, it must also go to 1.

Therefore \displaystyle \displaystyle \begin{align*} \lim_{ h \to 0} \frac{\sin{(h)}}{h} = 1 \end{align*}

Note: Technically we have only proven the right hand limit, where \displaystyle \displaystyle \begin{align*} \theta \to 0 \end{align*} from the positive direction, but the proof is almost identical for the left hand limit, you just need to work in the fourth quadrant and be careful with negatives and the direction the inequality signs point.
• Jun 28th 2013, 08:39 AM
math604
Re: Finding the limit
Great! I understand it now.

Thanks for the really long explanation. After I read it, I realized the my next step was that the limit of -sinh/h was 1, since sinh/h equaled 1.

Thanks!
• Jun 28th 2013, 07:28 PM
Prove It
Re: Finding the limit
Actually your limit would be -1.