Limit of a trigonometric function

[Jimmy84]

Homework Statement

Find the limit as x approaches 0 of

f(x) = 3x / sin 5x

Homework Equations

I might have to use the fact that "the lim of x approaches 0 of sin x/x = 1"

The Attempt at a Solution

I don't know how to start to solve the expresion, I would appreciate any help.
thanks.

 

[Dick]

Substitute u=5x. So u also approaches 0. Then use lim u/sin(u)=1 as u->0.

 

[rl.bhat]

Homework Statement

Find the limit as x approaches 0 of

f(x) = 3x / sin 5x

You can rewrite f(x) as
f(x) = 3/5*5x/sin5x and the find the limit.

 

[n!kofeyn]

A good explanation of this problem lies in this https://www.physicsforums.com/showthread.php?t=305154".

I think the best advice is to combine the above two suggestions. Rewrite the limit as
$$\lim_{x\to0} \frac{3x}{\sin 5x} = \lim_{x\to0} \frac{3}{5} \frac{5x}{\sin 5x} = \frac{3}{5}\lim_{x\to0} \frac{5x}{\sin 5x}$$

Now let u=5x, and follow the example shown in the posted link.

 

[Jimmy84]

thanks a lot guys, the explanation you gave in the other post were useful for me, my book didnt explain that way of substitution a lot.


right now I am having a difficult time subtituting this:

1.) f(x) 3x^2 / 1 - cos 1/2 x^2

I think that for this problem I got to use the fact that x - cos x / x = 0


and
2.) g(x) sin x /x^2


at first sight, these problems arent substutited in the same was as you described in the previous post though. Sometimes when I try to substitute terms I get all confused and end up mading up answers for the trigonometric functions, I was wondering if there are certain substitution steps that I could use for most of the trigonometric functions? like the one you described in the previous posts.

take care and thanks a lot.

 

[rock.freak667]

1.) f(x) 3x^2 / 1 - cos 1/2 x^2

I think that for this problem I got to use the fact that x - cos x / x = 0

is this supposed to be

$$\lim_{x \rightarrow 0} \frac{3x^2}{1-cos\frac{x^2}{2}}$$

2.) g(x) sin x /x^2

$$\lim_{x \rightarrow 0} \frac{sinx}{x^2}$$


if that is the question then try using the fact that 1/x² = (1/x)*(1/x)
and

$$\lim_{x \rightarrow a} f(x)g(x) = \lim_{x \rightarrow a}f(x) * \lim_{x \rightarrow a} g(x)$$

 

[Jimmy84]

is this supposed to be

$$\lim_{x \rightarrow 0} \frac{3x^2}{1-cos\frac{x^2}{2}}$$




$$\lim_{x \rightarrow 0} \frac{sinx}{x^2}$$


if that is the question then try using the fact that 1/x² = (1/x)*(1/x)
and

$$\lim_{x \rightarrow a} f(x)g(x) = \lim_{x \rightarrow a}f(x) * \lim_{x \rightarrow a} g(x)$$

the second problem is right, the first one is 3x^2 / (1-cos 1/2 x)^2
sorry I am having problems with the notation.

 

[HallsofIvy]

Then both multiply and divide the numerator by 1/4 to get
[tex]\frac{12\frac{x^2}{4}}{\left(1- cos \frac{x}{2}\right)^2}= 12\left(\frac{\frac{x}{2}}{1- cos(\frac{x}{2})}\right)^2[/itex]
and use the fact that
[tex]\lim_{a\rightarrow 0}\frac{1- cos(u)}{u}= 0[/itex]
But note that your fraction is the reciprocal of that- that's very important here!