Solving Limits: Seeking Help with x Approaching 0

[vgower]

Limits

I'm doing some online practice work... and I can't find out how to get the following two limits... I know they are not undefined.

The limit as x approaches 0 of sin^2(4x) / xtan(x) .. I tried to flip it... but I couldn't get it that way either.


The limit as x approaches 0 of (x + tan(4x)) / sin(9x)

Any help would be appreciated.

 

[dmoravec]

have you tried L'Hospital's rule yet?
http://mathworld.wolfram.com/LHospitalsRule.html

 

[vgower]

We're not doing derivitaves yet.. so it wouldn't make sense.. but I could try.

 

[vgower]

Doesn't work.

 

[vgower]

I'm doing some online practice work... and I can't find out how to get the following two limits... I know they are not undefined.

The limit as x approaches 0 of sin^2(4x) / xtan(x) .. I tried to flip it... but I couldn't get it that way either.


The limit as x approaches 0 of (x + tan(4x)) / sin(9x)

Any help would be appreciated.

 

[Swapnil]

I am getting 16 as my answer. I first broke \sin(4x)^2 into (2\sin(2x)*\cos(2x))^2 and then I further broke this down to 16\sin(x)^2\cos(x)^2(\cos(x)^2-\sin(x)^2)^2. Then I didived it by x*tan(x) from which I got:

16*\frac{\sin(x)}{x}\cos(x)^3(\cos(x)^2-\sin(x)^2)

Then ... here is the muddy part ... I assumed that the limit of the products is the product of the limits. Then, since the limit of sinx(x)/x is 1, you end up with 16 as your answer after you evalute the rest of the limit at x=0.

 

[Hurkyl]

Then ... here is the muddy part ... I assumed that the limit of the products is the product of the limits.

Then find a reason why you can assume that. It's not hard.


Incidentally, if you know (sin x)/x --> 1 as x --> 0, there's a much easier way to deal with sin (4x)...

 

[dmoravec]

i also got 16 for the first one using L'Hopitals rule (although I had to go to second derivatives) but I just expanded sin^2(4x) as [1-cos(8x)]/2 and did l'hopitals rule twice (making sure I kept my fractions seperate)

 

[Hurkyl]

You're all making these way too hard. Think about it -- what is the intuitive meaning of:

<br /> \lim_{x \rightarrow 0} \frac{\sin x}{x}<br />

? Does it tell you anything interesting geometrically or algebraically?

Highlight if you need to see the answer: (it tells you that, when x is small, sin x looks and acts very much like x[/color])

So what does that tell you about these limits?

 

[HallsofIvy]

I'm doing some online practice work... and I can't find out how to get the following two limits... I know they are not undefined.

The limit as x approaches 0 of sin^2(4x) / xtan(x) .. I tried to flip it... but I couldn't get it that way either.

The simplest method would be L'Hopital's rule. Have you tried that? If you don't want to do that, you might separate it as
4\frac{sin(4x)}{4x}\frac{sin(4x)}{tan(x)}
You should know the limit of 4\frac{sin(4x)}{4x} and you should be able to use the trig identity sin(4x)= 2sin(2x)cos(2x) to reduce
\frac{sin(4x)}{tan(x)}

 

[HallsofIvy]

This same question was asked under "homework". I am going to combine the two threads.

 

[Hurkyl]

you should be able to use the trig identity sin(4x)= 2sin(2x)cos(2x) to reduce

You shouldn't need the trig identity. The only facts you need are the elementary limit properties, and that the limits of (sin x)/x and (tan x)/x are both 1, as x goes to zero.