trig limit question

[dnt]

1. The problem statement, all variables and given/known data

evaluate the limit as x goes to 0 of:

tan(4x^2)^2/(3x^4)

(thats suppose to say the tangent squared of 4 x squared all over 3 x to the 4th)

2. Relevant equations

n/a

3. The attempt at a solution

the only trig limit stuff i know is the limit to 0 of sinx / x = 1

and

limit to 0 of (1-cos x)/x = 0

is there a tangant one i should know? what about tangant squared? i also considered changing it to sin squared over cos squared but i dont know what i could do that with.

if anyone could give me a hint to get me started, i would appreciate it. thanks.

[dnt]

hmmm actually i think i have an idea (strange how just typing it out can make your mind think in different ways).

after changing it to sin^2 over cos^2 i can then rewrite it like this:

(1/3)(sin4x^2/x^2)(sin4x^2/x^2)(1/cos4x^2)(1/cos4x^2)

then multiply the two sin parts by 4/4 which if i then apply the limit will give me:

(1/3)(4/1)(4/1)(1/1)(1/1)

which is 16/3

is that correct?

[berkeman]

tan(4x^2)^2/(3x^4)

(thats suppose to say the tangent squared of 4 x squared all over 3 x to the 4th)


hmmm actually i think i have an idea (strange how just typing it out can make your mind think in different ways).

after changing it to sin^2 over cos^2 i can then rewrite it like this:

(1/3)(sin4x^2/x^2)(sin4x^2/x^2)(1/cos4x^2)(1/cos4x^2)

then multiply the two sin parts by 4/4 which if i then apply the limit will give me:

(1/3)(4/1)(4/1)(1/1)(1/1)

which is 16/3

is that correct?

In LaTex:

\frac{tan^2 (4x^2)}{3x^4}

or do you mean:

\frac{tan^2 (4x)^2}{(3x)^4}

Or something else? If you quote my reply, you will see the format of the LaTex inside... It would help to see your equations in LaTex to eliminate the confusion. There's also a new LaTex editing feature in the reply dialog -- in the upper right corner of the entry box, there is a \Sigma symbol to click to get the LaTex editor.

[dnt]

its this one:

\frac{tan^2 (4x^2)}{3x^4}

[berkeman]

its this one:

\frac{tan^2 (4x^2)}{3x^4}

Okay. Now can you show your steps to the solution using LaTex to make them easier to check? Thanks.

BTW, the LaTex preview feature now works here on the PF (as of a day or two ago). When you're in the Advanced Reply window, just click on "Preview Post" to make sure you're happy with the appearance of the LaTex.

[Dick]

You've got two ways to go here. You can blindly apply L'Hopital's rule through 4 differentiations or you can look at an expansion of tan(x) for small x and go more or less directly to the answer.

[chaoseverlasting]

Or you can write it as (\frac{tan4x^2}{4x^2})^2\frac{16}{3}. Can you find the limit now?

[VietDao29]

hmmm actually i think i have an idea (strange how just typing it out can make your mind think in different ways).

after changing it to sin^2 over cos^2 i can then rewrite it like this:

(1/3)(sin4x^2/x^2)(sin4x^2/x^2)(1/cos4x^2)(1/cos4x^2)

then multiply the two sin parts by 4/4 which if i then apply the limit will give me:

(1/3)(4/1)(4/1)(1/1)(1/1)

which is 16/3

is that correct?

Yes, that's correct. :smile:

When seeing tangent function, you can just split it into sin, and cos function, and simply go from there.

[dnt]

Or you can write it as (\frac{tan4x^2}{4x^2})^2\frac{16}{3}. Can you find the limit now?


interesting. since i know the answer is 16/3, can i assume the limit of (\frac{tan4x^2}{4x^2})^2\ when approaching 0 is 1? ive never heard of that limit rule for tangent before.

[VietDao29]

interesting. since i know the answer is 16/3, can i assume the limit of (\frac{tan4x^2}{4x^2})^2\ when approaching 0 is 1? ive never heard of that limit rule for tangent before.

Well, actually, you can derive it from the well-known limit:
\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1

Just split tan into sin, and cos, like this:
\lim_{x \rightarrow 0} \frac{\tan x}{x} = \lim_{x \rightarrow 0} \frac{\sin x}{x} \frac{1}{\cos x} = 1 \times \frac{1}{1} = 1

[chaoseverlasting]

Yeah... thats it.