# Math Help - find limit as x approach 0 : [cos(2x) -1]/sin7x

1. ## find limit as x approach 0 : [cos(2x) -1]/sin7x

find limit as x approach 0 : [cos(2x) -1]/sin7x

Can anyone explain how I can start this problem? I am clueless!

2. ## Re: find limit as x approach 0 : [cos(2x) -1]/sin7x

You can use l'Hopital's rule.

3. ## Re: find limit as x approach 0 : [cos(2x) -1]/sin7x

Originally Posted by NeoSonata
find limit as x approach 0 : [cos(2x) -1]/sin7x

Can anyone explain how I can start this problem? I am clueless!
Hi NeoSonata,

Since both the numerator and denominator tends to zero as x tends to zero, this is an indeterminate form. I suggest you use the L'Hôpital's rule.

4. ## Re: find limit as x approach 0 : [cos(2x) -1]/sin7x

[(-2sin2x)(sin7x)-(7cos7x)(cos2x-1)]/(sin7x)^2

Is this how it works? I know how to use product and quotient rule but I am unfamiliar with dealing with "sin7x", "cos2x-1", anything similar to that.

5. ## Re: find limit as x approach 0 : [cos(2x) -1]/sin7x

That's not what L'Hopitals rule says, you have to take the derivative of the numerator and the derivative of the denominator, not of the fraction. Try it again.

6. ## Re: find limit as x approach 0 : [cos(2x) -1]/sin7x

derivative of the numerator and the derivative of the denominator :

-2sin2x / 7cos7x

7. ## Re: find limit as x approach 0 : [cos(2x) -1]/sin7x

Correct. Can you determine the limit now? ...

8. ## Re: find limit as x approach 0 : [cos(2x) -1]/sin7x

Originally Posted by Sudharaka
Hi NeoSonata,

Since both the numerator and denominator tends to zero as x tends to zero, this is an indeterminate form. I suggest you use the L'Hôpital's rule.
The power series for cos(2x) and sin(7x) can also be substituted, the result simplified and then the limit easily calculated.

9. ## Re: find limit as x approach 0 : [cos(2x) -1]/sin7x

Originally Posted by NeoSonata
derivative of the numerator and the derivative of the denominator :

-2sin2x / 7cos7x
Yes that is correct. I think you have to understand the L'Hôpital's rule. So I will briefly explain. Suppose you have to evaluate a limit of the form, $\lim_{x\rightarrow c}\frac{f(x)}{g(x)}=\frac{0}{0}\mbox{ or }\lim_{x\rightarrow c}\frac{f(x)}{g(x)}=\frac{\pm\infty}{\pm\infty}$ where f(x) and g(x) are two differentiable functions and $c\in\Re\mbox{ or }c=\pm\infty$

Then the L'Hôpital's rule says that, $\lim_{x\rightarrow c}\frac{f(x)}{g(x)}=\lim_{x\rightarrow c}\frac{f'(x)}{g'(x)}$

So in this case you have to evaluate, $\lim_{x\rightarrow 0}\frac{\cos 2x-1}{\sin7x}=\lim_{x\rightarrow 0}\frac{-2\sin 2x}{7\cos 7x}=0$

A more detailed (and easily understandable) description about the L'Hôpital's rule could be found at, Pauls Online Notes : Calculus I - L'Hospital's Rule and Indeterminate Forms

10. ## Re: find limit as x approach 0 : [cos(2x) -1]/sin7x

I would consider both "L'Hopital's rule" and "power series" to be 'overkill' for a problem like this. Write the fraction as
$\frac{cos(2x)- 1}{sin(7x)}= \left(\frac{2}{7}\right)\left(-\frac{1- cos(2x)}{2x}\right)\left(\frac{7x}{sin(7x)}\right)$

And then use $\lim_{u\to 0}\frac{sin(u)}{u}= 1$ and $\lim_{v\to 0}\frac{1- cos(v)}{v}= 0$

11. ## Re: find limit as x approach 0 : [cos(2x) -1]/sin7x

Hello, NeoSonata!

From the "shape" of the problem, I suspect these theorems are suggested:

. . $\lim_{\theta\to0}\frac{\cos\theta - 1}{\theta} \;=\;0 \qquad\quad \lim_{\theta\to0}\frac{\sin\theta}{\theta} \;=\;1$

$\lim_{x\to0}\frac{\cos(2x) -1}{\sin(7x)}$

We have: . $\frac{\cos(2x)-1}{1}\cdot \frac{1}{\sin(7x)}$

Multiply the first fraction by $\tfrac{2x}{2x}$, the second by $\tfrac{7x}{7x}$

. . $\displaystyle \frac{2x}{2x}\cdot\frac{\cos(2x)-1}{1}\cdot\frac{7x}{7x}\cdot\frac{1}{\sin(7x)} \;=\;\frac{2}{7}\cdot\frac{\cos(2x)-1}{2x}\cdot\frac{7x}{\sin(7x)}$

$\text{When }x\to0,\,\text{ both }2x\to 0 \text{ and }7x\to0.$

$\text{We have: }\:\lim_{x\to0}\left[\frac{2}{7}\cdot\frac{\cos(2x)-1}{2x}\cdot\frac{7x}{\sin(7x)}\right]$

. . . . . $=\;\frac{2}{7}\left[\lim_{2x\to0}\frac{\cos(2x)-1}{2x}\right]\,\left[\lim_{7x\to0}\frac{7x}{\sin(7x)}\right]$

. . . . . $=\;\frac{2}{7}\cdot 0 \cdot 1 \;\;=\;\;0$

Edit: Ah, HallsofIvy beat me to it! . . . *sigh*