Finding the Limit of 4t2*(sin(2/t))2

[Jude075]

Homework Statement

Lim (4t²)*(sin(2/t))²
t→∞

Homework Equations

The Attempt at a Solution

I know it will be a indeterminant form (∞*0) if I don't do anything to the expression, but I don't how.

 

[vanhees71]

Taylor expansion of the sine helps!

 

[Jude075]

Taylor expansion of the sine helps!

I am in Cal I

 

[Infrared]

Hint: lim_{t \rightarrow \infty} t \sin(1/t)= lim_{x \rightarrow 0} \frac{sin(x)}{x}=1 by making the substitution x=1/t. Use this combined with the double angle formula for sine to compute your limit.

 

[Jude075]

Hint: lim_{t \rightarrow \infty} t \sin(1/t)= lim_{x \rightarrow 0} \frac{sin(x)}{x}=1 by making the substitution x=1/t. Use this combined with the double angle formula for sine to compute your limit.

Why did you change t→∞ to x→0? I don't understand this.

 

[Infrared]

If x=1/t then x goes to 0 as t becomes large.

 

[micromass]

Have you seen L'Hospitals rule?

 

[Jude075]

I still can't do it :(

 

[Dick]

I still can't do it :(

Probably because you didn't pay much attention to HS-Scientist's fine suggestion. Try it. Change the variable to x where x=1/t. So if t->infinity then x->0.

 

[Jude075]

Okay. Here is how I tried.
Lim (4*(sin x)²)/x² because no much difference between 2/t and 1/t
X→0

Then I got an indeterminant form 0/0, so I used L'hopital's rule.
Lim (8sin x cos x)/(2x)
X→0

Still 0/0 ,so again
Lim (8(cosx)²-8(sinx)²)/2
X→0
Which equals 4.
But my calculator says the answer should be 16

 

[Mark44]

Okay. Here is how I tried.
Lim (4*(sin x)²)/x² because no much difference between 2/t and 1/t
X→0

Both 2/t and 1/t approach 0 as t gets large, but you can't ignore the difference.

In your original limit, you had sin²(2/t), so what does that become with the substitution x = 1/t?

Then I got an indeterminant form 0/0, so I used L'hopital's rule.
Lim (8sin x cos x)/(2x)
X→0

Still 0/0 ,so again
Lim (8(cosx)²-8(sinx)²)/2
X→0
Which equals 4.
But my calculator says the answer should be 16

 

[Jude075]

Both 2/t and 1/t approach 0 as t gets large, but you can't ignore the difference.

In your original limit, you had sin²(2/t), so what does that become with the substitution x = 1/t?

Right! That is what made me wrong. Thank you so much!
But I got the answer without using double-angle formula. Just keep applying L'hopital's rule:)

 

[Mark44]

You don't need either the double-angle formula or L'Hopital's Rule if you use this limit:
$$\lim_{x \to 0} \frac{\sin(x)}{x} = 1$$
This was suggested by HS-Scientist back in post #4.

 

[Jude075]

You don't need either the double-angle formula or L'Hopital's Rule if you use this limit:
$$\lim_{x \to 0} \frac{\sin(x)}{x} = 1$$
This was suggested by HS-Scientist back in post #4.

$$\lim_{x \to 0} \frac{4\sin^2(2x)}{x^2}$$

How do you use that limit to get 16?
This question comes from the AP test. So if I were to use the way I did to solve it , it would waste lots of time.

 

[Infrared]

$$\lim_{x \to 0} \frac{4\sin^2(2x)}{x^2}$$

How do you use that limit to get 16?
This question comes from the AP test. So if I were to use the way I did to solve it , it would waste lots of time.

$$\lim_{x \to 0} \frac{4\sin^2(2x)}{x^2}=4 (\lim_{x \to 0} \frac{sin(2x)}{x})^2=4(2^2)=16 $$

Believe me, you would have enough time on the AP test to do this. I took it just this year.

 

[Jude075]

$$\lim_{x \to 0} \frac{4\sin^2(2x)}{x^2}=4 (\lim_{x \to 0} \frac{sin(2x)}{x})^2=4(2^2)=16 $$

Believe me, you would have enough time on the AP test to do this. I took it just this year.

I didn't know you could just factor 2 out of the sin. Is there any properties or laws for this?

 

[Infrared]

I didn't know you could just factor 2 out of the sin. Is there any properties or laws for this?

There are a few ways you can think about \lim_{x \rightarrow 0} \frac{sin(2x)}{x}. You can use the double angle formula for sine to write sin(2x)=2sin(x)cos(x), which makes the limit clear, but this doesn't generalize very nicely to \lim_{x \rightarrow 0} \frac{sin(kx)}{x} where k is some other constant.

Instead, you can write \lim_{x \to 0} \frac{sin(2x)}{x}=2 \lim_{x \to 0} \frac{sin(2x)}{2x}=2 \lim_{2x \to 0} \frac{sin(2x)}{2x}=2

Or, you can make the substitution u=2x, which turns the limit into \lim_{u \rightarrow 0} \frac{sin(u)}{(u/2)}=2. This suggests that a better original substitution would have been x=\frac{2}{t}

Of course, you could have just used L'Hopital's Rule on sin(2x)/x but there is no reason to rely on derivatives for this limit.

tldr: \lim_{x \to 0} \frac{sin(kx)}{x}=k

 

[Jude075]

There are a few ways you can think about \lim_{x \rightarrow 0} \frac{sin(2x)}{x}. You can use the double angle formula for sine to write sin(2x)=2sin(x)cos(x), which makes the limit clear, but this doesn't generalize very nicely to \lim_{x \rightarrow 0} \frac{sin(kx)}{x} where k is some other constant.

Instead, you can write \lim_{x \to 0} \frac{sin(2x)}{x}=2 \lim_{x \to 0} \frac{sin(2x)}{2x}=2 \lim_{2x \to 0} \frac{sin(2x)}{2x}=2

Or, you can make the substitution u=2x, which turns the limit into \lim_{u \rightarrow 0} \frac{sin(u)}{(u/2)}=2. This suggests that a better original substitution would have been x=\frac{2}{t}

Of course, you could have just used L'Hopital's Rule on sin(2x)/x but there is no reason to rely on derivatives for this limit.

tldr: \lim_{x \to 0} \frac{sin(kx)}{x}=k

Good explanation! Thank you very much!