Finding a trigonometric limit

1. Sep 20, 2009

MassInertia

This is from Thomas' Calculus Early Trancendentals, Media Upgrade, page 108, #27:

The limit as x approaches 0 of (x csc(2x))/cos(5x)

The answer is 1/2, which graphing confirms, but hell if I know how to get rid of the 5x and still come up with 1/2. It is boggling my mind. Any help would be appreciated.

2. Sep 20, 2009

rock.freak667

csc(2x)=1/sin(2x)

if I remember correctly

$$\lim_{x \rightarrow 0} \frac{x}{sinx} =1$$

3. Sep 20, 2009

MassInertia

Yes, that much I know. The problem I have is with the cos(5x). How does one deal with that and come up with 1/2 as the final answer?

4. Sep 20, 2009

Gregg

Re-arrange then use L'Hopital

$$\displaystyle\lim_{x\to 0} \frac{x\csc(2x)}{\cos(5x)}$$

$$= \displaystyle\lim_{x\to 0} \frac{x\csc(2x)}{1}$$

$$= \displaystyle\lim_{x\to 0} \frac{x}{\sin(2x)} \to \frac{0}{0}$$

$$= \displaystyle\lim_{x\to 0} \frac{1}{2\cos(2x)} = \frac{1}{2}$$

Last edited: Sep 20, 2009
5. Sep 20, 2009

MassInertia

Well, we haven't covered L'Hopital yet. Also, the denominator is cos(5x), not 5x.

6. Sep 20, 2009

Bohrok

You need to rewrite the limit to use other limits that you know, such as sinx/x

$$\lim_{x \rightarrow 0} \frac{x\csc2x}{\cos5x} = \lim_{x \rightarrow 0} (x\csc2x * \frac{1}{\cos5x}) = \lim_{x \rightarrow 0} \frac{x}{\sin2x} * \lim_{x \rightarrow 0} \frac{1}{\cos5x}$$

$$= \lim_{x \rightarrow 0} \frac{1}{2} \frac{2x}{\sin2x} * \lim_{x \rightarrow 0} \frac{1}{\cos5x} = \frac{1}{2} \lim_{x \rightarrow 0} \frac{1}{\frac{\sin2x}{2x}} * \lim_{x \rightarrow 0} \frac{1}{\cos5x}$$

For $$\lim_{x \rightarrow 0} \frac{sin2x}{2x}$$

let k = 2x. Since 2x goes to 0 as k goes to 0,

$$\lim_{x \rightarrow 0} \frac{\sin2x}{2x} = \lim_{k \rightarrow 0} \frac{\sin k}{k}$$

Last edited: Sep 20, 2009
7. Sep 20, 2009

Gregg

I made a typo in latex, if it were 5x then we'd be in trouble.

L'Hopital

$$\displaystyle\lim_{x\to a} \frac{f(x)}{g(x)} = \displaystyle\lim_{x\to a}\frac{f'(x)}{g'(x)}$$

if f(a)=g(a)=0 (or infinity)

$$\frac{f(x)}{g(x)}=\frac{f(a)+(x-a)f'(a)+\frac{1}{2!}(x-a)^2g''(a)\cdots}{g(a)+(x-a)g'(a)+\frac{1}{2!}(x-a)^2g''(a)+\cdots}=\frac{(x-a)f'(a)+\frac{1}{2!}(x-a)^2g''(a)\cdots}{(x-a)g'(a)+\frac{1}{2!}(x-a)^2g''(a)+\cdots}=\frac{f'(a)+(x-a)g''(a)\cdots}{g'(a)+(x-a)g''(a)+\cdots}=\frac{f'(x)}{g'(x)}$$

f(a) and g(a) disappear and can divide top and bottom by 1/2(x-a).

example

$$\displaystyle\lim_{x\to 0} \frac{\sin (x)}{x}$$

Top and bottom tend to 0, so,

$$\displaystyle\lim_{x\to 0} \frac{\sin (x)}{x} =\displaystyle\lim_{x\to 0} \frac{\cos (x)}{1} = 1$$

8. Sep 20, 2009

Bohrok

I've read in my calc book and elsewhere that you shouldn't use L'Hopital's rule with sinx/x since that limit is used in proofs to find the derivative of sine. Circular reasoning...

9. Sep 20, 2009

MassInertia

Thanks, Bohrok, that's exactly what I needed to set my brain straight.

Perhaps if I would have remembered that the cosine of zero is one, I wouldn't have been in such a pickle. Hah.

Thanks to everyone for their assistance.