Calculating Limit: sin^2(3t)/t^2 as t → 0

  • Thread starter Taryn
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In summary: Thank you so much!In summary, the conversation is discussing how to calculate the limit lim sin^2(3t)/t^2 as t approaches 0. They mention using the limit \lim_{x\rightarrow 0}\frac{sin(x)}{x} = 1 and applying L'Hospital's rule to solve it. The conversation ends with the expert summarizer providing an example and asking if there are any remaining questions.
  • #1
Taryn
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Okay so this is the problem
Calculate this limit
lim sin^2(3t)/t^2
t->0

This is wat I did...

I sed that it is also the lim sin3t*sin3t/t*t
t->0
I know that sint is upper and lower bounded by -1 and 1
Now this is where I just get completely stuffed... and confused!

could someone just tell me or even show me how to think about doing the next step, and also tell me about the L'Hosp... rule... never been shown that and from other posts I've read it seems to make everythin a lot easier.
 
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  • #2
for l'hopital's rule, you take the limit of the top derivative over the bottom derivative.
 
  • #3
For your question, what is
[tex]\lim_{x\rightarrow 0}\frac{sin(x)}{x} ? [/tex]

L'Hospital rule is applicable only in cases where the expression whose limit is to be taken gives an indeterminate form when evaluated directly(functional value 0/0 or inf./inf.)
 
  • #4
arunbg said:
L'Hospital rule is applicable only in cases where the expression whose limit is to be taken gives an indeterminate form when evaluated directly(functional value 0/0 or inf./inf.)
And, in this case, it's the indeterminate form 0 / 0.
Taryn said:
Okay so this is the problem
Calculate this limit
lim sin^2(3t)/t^2
t->0

This is wat I did...

I sed that it is also the lim sin3t*sin3t/t*t
Uhmm, you know the well-known limit:
[tex]\lim_{t \rightarrow 0} \frac{\sin t}{t} = 1[/tex], right?
That limit can be applied nicely here. I'll give you an example:
Example:
Evaluate
[tex]\lim_{u \rightarrow 0} \frac{\sin (2u)}{u}[/tex]
--------------
Since there is 2u in the numerator, while in the denominator there's only u, we'll try to multiply both numerator, and denominator by 2 to get:
[tex]\lim_{u \rightarrow 0} \frac{\sin (2u)}{u} = \lim_{u \rightarrow 0} \frac{2 \sin (2u)}{2u} = 2 \lim_{u \rightarrow 0} \frac{\sin (2u)}{2u}[/tex]
Now, as u tends to 0, 2u also tends to 0, right? If we let y = 2u, then as u -> 0, y also tends to 0, right?
So:
[tex]\lim_{u \rightarrow 0} \frac{\sin (2u)}{u} = 2 \lim_{y \rightarrow 0} \frac{\sin (y)}{y} = 2 \times 1 = 2[/tex].
Is there anything unclear?
Can you go from here? :)
 
Last edited:
  • #5
yeah cheers for that I should be able to figure it out from there
 

1. What is the limit of sin^2(3t)/t^2 as t approaches 0?

The limit of sin^2(3t)/t^2 as t approaches 0 is equal to 3.

2. How do you calculate the limit of a function as it approaches 0?

To calculate the limit of a function as it approaches 0, you can simply plug in 0 for the variable and solve the resulting expression.

3. What does it mean when a limit approaches 0?

When a limit approaches 0, it means that the values of the function are getting closer and closer to 0 as the input values approach 0.

4. Can the limit of a function be undefined as it approaches 0?

Yes, the limit of a function can be undefined as it approaches 0 if the numerator of the function becomes 0 and the denominator becomes 0 as well.

5. How can you use L'Hopital's rule to calculate the limit of a function as it approaches 0?

L'Hopital's rule states that if the limit of a function as it approaches a certain value is indeterminate (such as 0/0), then you can take the derivative of the numerator and denominator separately and evaluate the limit using the resulting expression. This can be applied to calculate the limit of a function as it approaches 0.

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