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.
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, $\displaystyle \lim_{x\rightarrow c}\frac{f(x)}{g(x)}=\frac{0}{0}\mbox{ or }\lim_{x\rightarrow c}\frac{f(x)}{g(x)}=\frac{\pm\infty}{\pm\infty}$ where f(x) and g(x) are two differentiable functions and $\displaystyle c\in\Re\mbox{ or }c=\pm\infty$
Then the L'Hôpital's rule says that, $\displaystyle \lim_{x\rightarrow c}\frac{f(x)}{g(x)}=\lim_{x\rightarrow c}\frac{f'(x)}{g'(x)}$
So in this case you have to evaluate, $\displaystyle \lim_{x\rightarrow 0}\frac{\cos 2x-1}{\sin7x}=\lim_{x\rightarrow 0}\frac{-2\sin 2x}{7\cos 7x}=0$
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
I would consider both "L'Hopital's rule" and "power series" to be 'overkill' for a problem like this. Write the fraction as
$\displaystyle \frac{cos(2x)- 1}{sin(7x)}= \left(\frac{2}{7}\right)\left(-\frac{1- cos(2x)}{2x}\right)\left(\frac{7x}{sin(7x)}\right)$
And then use $\displaystyle \lim_{u\to 0}\frac{sin(u)}{u}= 1$ and $\displaystyle \lim_{v\to 0}\frac{1- cos(v)}{v}= 0$
Hello, NeoSonata!
From the "shape" of the problem, I suspect these theorems are suggested:
. . $\displaystyle \lim_{\theta\to0}\frac{\cos\theta - 1}{\theta} \;=\;0 \qquad\quad \lim_{\theta\to0}\frac{\sin\theta}{\theta} \;=\;1$
$\displaystyle \lim_{x\to0}\frac{\cos(2x) -1}{\sin(7x)} $
We have: .$\displaystyle \frac{\cos(2x)-1}{1}\cdot \frac{1}{\sin(7x)} $
Multiply the first fraction by $\displaystyle \tfrac{2x}{2x}$, the second by $\displaystyle \tfrac{7x}{7x}$
. . $\displaystyle \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)} $
$\displaystyle \text{When }x\to0,\,\text{ both }2x\to 0 \text{ and }7x\to0.$
$\displaystyle \text{We have: }\:\lim_{x\to0}\left[\frac{2}{7}\cdot\frac{\cos(2x)-1}{2x}\cdot\frac{7x}{\sin(7x)}\right] $
. . . . . $\displaystyle =\;\frac{2}{7}\left[\lim_{2x\to0}\frac{\cos(2x)-1}{2x}\right]\,\left[\lim_{7x\to0}\frac{7x}{\sin(7x)}\right] $
. . . . . $\displaystyle =\;\frac{2}{7}\cdot 0 \cdot 1 \;\;=\;\;0$
Edit: Ah, HallsofIvy beat me to it! . . . *sigh*