2

I am taking an online course in Calculus from Ohio State and am just being introduced to the concept of limits. One of the exercises given to me is to find the limit of the following

$$\lim_{x \rightarrow 0}\frac {\sin({2x})}{8x} $$ Using what I have so far been taught, I determined that this is equivalent to the following

$$\frac {\lim_{x \rightarrow 0}\sin(2x)} {(\lim_{x \rightarrow 0}8) (\lim_{x \rightarrow 0}x)} $$ As far as I am aware, this should become

$$\frac 0 {(8)(0)}$$ which is undefined, so I have a feeling this is incorrect. I'm sure that this is an easy problem to solve, I'm just not sure how to do it without getting $\frac 0 0$ as an answer.

Martin Sleziak
  • 53,687
Chris Loonam
  • 123
  • 3
    Use the fact that $\lim\limits_{t\to0}\frac {\sin t} t=0$. Let $t=2x$. – John Molokach Nov 26 '15 at 00:43
  • One way to do this is to recall that $\lim_{t\to 0}\frac{\sin t}{t}=1$ so you can write your limit as $\frac{1}{4}\lim_{x\to 0}\frac{\sin 2x}{2x}=\frac{1}{4}$. – Jack Nov 26 '15 at 00:44

5 Answers5

4

$\sin(x) = x$ as $x$ approaches $0$, therefore $\frac{\sin(2x)}{8x} =\frac{2x}{8x} = \frac14$.

Ori
  • 398
  • 1
  • 3
  • 13
Robert S
  • 1,144
3

Here's a case where you use L'Hôpital's rule.

Given two functions $f(x)$ and $g(x)$, if $$\lim_{x\to c}\frac{f(x)}{g(x)}$$ is indeterminate (for all intents and purposes, undefined in this case), then $$\lim_{x\to c}\frac{f(x)}{g(x)}=\lim_{x\to c}\frac{f'(x)}{g'(x)}$$ and the same for higher derivatives.

Therefore, in this case, all you have to do is take the derivative of the top and the derivative of the bottom, and find the limit as $x\to0$. $$\lim_{x\to0}\frac{\sin(2x)}{8x}=\lim_{x\to0}\frac{\frac{\mathrm{d}}{\mathrm{d}x}(\sin(2x))}{\frac{\mathrm{d}}{\mathrm{d}x}(8x)}$$ Can you simplify from here?

HDE 226868
  • 2,354
3

THIS QUESTION'S "BEST" ANSWER, uses basic tools only to establish the inequalities for the sine function

$$\cos z\le \frac{\sin z}{z}\le 1 \tag 1$$

for $-\pi/2 \le z\le \pi/2$.

Then, letting $z=2x$ in $(1)$ reveals

$$\cos (2x)\le \frac{\sin(2x)}{2x}\le 1 \tag 2$$

for $-\pi/4\le x\le \pi/4$.

Dividing both sides of $(2)$ by $4$ we obtain

$$\frac14 \cos (2x)\le \frac{\sin(2x)}{8x}\le \frac14 $$

whereupon applying the Squeeze Theorem yields

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0}\frac{\sin(2x)}{8x}=\frac14}$$

as expected!

Community
  • 1
Mark Viola
  • 179,405
2

If you have no prior experience with derivatives, you might want to try substituting the identity $\sin{2x} = \sin{x}\cos{x}$

user293517
  • 19
2

Although L'Hopital's rule can be used, this can be done using the fundamental limit \begin{equation} \lim_{x \to 0} \frac{\sin x}{x} = 1 \end{equation} This can be proven using the unit circle(try graphing the function to see why this is true). Now remember that as $x \to 0$, $2x \to 0$. So allowing $y = 2x$ the limit becomes \begin{equation} \lim_{y\to 0} \frac{\sin y}{4y} = \frac{1}{4} \lim_{y\to 0} \frac{\sin y}{y} = \frac{1}{4} \cdot 1 = \frac{1}{4} \end{equation}

Jeevan Devaranjan
  • 2,575