Trig limits

**Foxlion** · April 30th 2012, 05:51 PM

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

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

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

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

**skeeter** · April 30th 2012, 06:20 PM

Originally Posted by Foxlion

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

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

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

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

you should already be familiar with this well known trig limit ...

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

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

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

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

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

**Soroban** · April 30th 2012, 06:29 PM

Hello, skeeter!

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

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

I'm afraid that he will ask:

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

**Foxlion** · April 30th 2012, 08:01 PM

Originally Posted by Soroban

Hello, skeeter!

I'm afraid that he will ask:

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

That's funny...

**Prove It** · April 30th 2012, 09:07 PM

Originally Posted by Soroban

Hello, skeeter!

I'm afraid that he will ask:

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

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

**Foxlion** · April 30th 2012, 09:48 PM

Originally Posted by Prove It

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

Exactly!, we'll call it a corollary prop of the alphabetical cancellation rule...

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