1. ## L'hospitals rule mistake?

lim x-> 0 of sin5x/sinx

Basically since its 0/0 i used L'hospitals rule and took the derivative of top and bottom to get:
5cos5x/cosx
which it then 5(1)/1... so =5?

But then i put it into this website that calculates it for you and it said the answer is 1 soo im a tad confused.

2. Originally Posted by kensington
lim x-> 0 of sin5x/sinx

Basically since its 0/0 i used L'hospitals rule and took the derivative of top and bottom to get:
5cos5x/cosx
which it then 5(1)/1... so =5?

But then i put it into this website that calculates it for you and it said the answer is 1 soo im a tad confused.
it's 5 where did you calculate ?

try here Wolfram|Alpha&mdash;Computational Knowledge Engine

3. $\displaystyle \displaystyle \lim_{x \to 0} \frac{\sin(5x)}{\sin{x}} = 5$

4. Originally Posted by kensington
lim x-> 0 of sin5x/sinx

Basically since its 0/0 i used L'hospitals rule and took the derivative of top and bottom to get:
5cos5x/cosx
which it then 5(1)/1... so =5?

But then i put it into this website that calculates it for you and it said the answer is 1 soo im a tad confused.
I beleive that your answer is correct, and that either the website made a mistake or you might've typed it in wrong (since using calculator syntax can be a cause for error in a problem like this one). I support your answer, not only in agreeing with your methodology, but I ploted $\displaystyle y =\frac{sin(5x)}{sin(x)}$ and it most certainly appears to approach 5 as $\displaystyle x \to 0$. I could be wrong though. Lets see what other people say.

5. Originally Posted by yeKciM
it's 5 where did you calculate ?

try here Wolfram|Alpha&mdash;Computational Knowledge Engine
I used Mathway: Math Problem Solver and i know i typed it in right because it showed me the question and then the answer.. wierd.
But i was ready to give up because i didnt see how they could have gotten 1

6. Originally Posted by kensington
I used Mathway: Math Problem Solver and i know i typed it in right because it showed me the question and then the answer.. wierd.
But i was ready to give up because i didnt see how they could have gotten 1
to try it in wolfram type

lim_(x->0) sin(5x)/sin(x)

many times (for "easy" calculating) you have steps if you need to check your work...

7. Originally Posted by kensington
lim x-> 0 of sin5x/sinx

Basically since its 0/0 i used L'hospitals rule and took the derivative of top and bottom to get:
5cos5x/cosx
which it then 5(1)/1... so =5?

But then i put it into this website that calculates it for you and it said the answer is 1 soo im a tad confused.
Alternatively...

$\displaystyle \displaystyle\huge\frac{ Sin(5x)}{Sin(x)}=\frac{5x}{5x}\frac{Sin(5x)}{Sin(x )}=5\frac{x}{Sin(x)}\frac{Sin(5x)}{5x}$

As "u" approaches zero, $\displaystyle Sinu\rightarrow\ u$

Now use $\displaystyle \displaystyle\huge\lim_{u\rightarrow0}\frac{Sinu}{ u}=1$

8. Originally Posted by Archie Meade
Alternatively...

$\displaystyle \displaystyle\huge\frac{ Sin(5x)}{Sin(x)}=\frac{5x}{5x}\frac{Sin(5x)}{Sin(x )}=5\frac{x}{Sin(x)}\frac{Sin(5x)}{5x}$

Now use $\displaystyle \displaystyle\huge\lim_{u\rightarrow0}Sinu=u$
Use instead $\displaystyle \lim_{u\to 0} \frac{sin u}{u}= 1$.

$\displaystyle \lim_{u\to 0} Sin u= 0$, not u.