# Thread: Trig limits

1. ## Trig limits

Find $\displaystyle \lim_{x\rightarrow1} \frac{sin(x-1)}{x^2+x-2}$

$\displaystyle = \lim_{x\rightarrow1}\frac{sin(x-1)}{(x+2)(x-1)}$

do the $\displaystyle (x-1)$ terms cancel out? If so, doesn't that leave me with $\displaystyle \lim_{x\rightarrow1}\frac{sinx}{x+2}$ ?

then by direct sub $\displaystyle \lim_{x\rightarrow1}\frac{sinx}{x+2} = \frac{sin1}{3}$ ?

2. ## Re: Trig limits

Originally Posted by Foxlion
Find $\displaystyle \lim_{x\rightarrow1} \frac{sin(x-1)}{x^2+x-2}$

$\displaystyle = \lim_{x\rightarrow1}\frac{sin(x-1)}{(x+2)(x-1)}$

do the $\displaystyle (x-1)$ terms cancel out? If so, doesn't that leave me with $\displaystyle \lim_{x\rightarrow1}\frac{sinx}{x+2}$ ?

then by direct sub $\displaystyle \lim_{x\rightarrow1}\frac{sinx}{x+2} = \frac{sin1}{3}$ ?
you should already be familiar with this well known trig limit ...

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

$\displaystyle \lim_{x \to 1} \frac{\sin(x-1)}{(x-1)(x+2)} =$

$\displaystyle \lim_{x \to 1} \frac{\sin(x-1)}{x-1} \cdot \frac{1}{x+2} =$

$\displaystyle \lim_{x \to 1} \frac{\sin(x-1)}{x-1} \cdot \lim_{x \to 1} \frac{1}{x+2} =$

$\displaystyle 1 \cdot \frac{1}{3} = \frac{1}{3}$

3. ## Re: Trig limits

Hello, skeeter!

You should already be familiar with this well known trig limit:

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

I'm afraid that he will ask:

. . Do the $\displaystyle \theta$ cancel out?
. . If so, doesn't that leave me with: .$\displaystyle \lim_{\theta\to0}\,\sin \:=\:1\,?$

4. ## Re: Trig limits

Originally Posted by Soroban
Hello, skeeter!

I'm afraid that he will ask:

. . Do the $\displaystyle \theta$ cancel out?
. . If so, doesn't that leave me with: .$\displaystyle \lim_{\theta\to0}\,\sin \:=\:1\,?$

That's funny...

5. ## Re: Trig limits

Originally Posted by Soroban
Hello, skeeter!

I'm afraid that he will ask:

. . Do the $\displaystyle \theta$ cancel out?
. . If so, doesn't that leave me with: .$\displaystyle \lim_{\theta\to0}\,\sin \:=\:1\,?$

Isn't this the same reason why \displaystyle \displaystyle \begin{align*} \frac{\sin{x}}{n} = \textrm{six} = 6 \end{align*}?

6. ## Re: Trig limits

Originally Posted by Prove It
Isn't this the same reason why \displaystyle \displaystyle \begin{align*} \frac{\sin{x}}{n} = \textrm{six} = 6 \end{align*}?
Exactly!, we'll call it a corollary prop of the alphabetical cancellation rule...