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

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Homework Help Overview

The problem involves calculating the limit of the expression sin²(3t)/t² as t approaches 0. Participants are exploring the application of L'Hôpital's rule and the known limit of sin(x)/x as x approaches 0.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to rewrite the limit expression and expresses confusion about the next steps. Some participants suggest using L'Hôpital's rule and reference the limit of sin(x)/x as a foundational concept.

Discussion Status

The discussion includes various interpretations of the limit problem, with some participants providing guidance on applying L'Hôpital's rule and the known limit of sin(x)/x. There is no explicit consensus, but a productive direction has been established.

Contextual Notes

Participants note the indeterminate form of 0/0 when evaluating the limit directly, which is relevant to the application of L'Hôpital's rule.

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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for l'hopital's rule, you take the limit of the top derivative over the bottom derivative.
 
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.)
 
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:
yeah cheers for that I should be able to figure it out from there
 

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