Unsolvable limit of trigonometric function

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To find the limit of sin(3t)^2/t^2 as t approaches 0 without using L'Hôpital's rule, one can utilize the known limit of sin(x)/x as x approaches 0. By rewriting sin(3t)^2/t^2 as (sin(3t)/t) * (sin(3t)/t) and applying the limit properties, it simplifies to 9 * (sin(3t)/(3t))^2. Since lim(t → 0) (sin(3t)/(3t)) equals 1, the overall limit evaluates to 9. This approach effectively resolves the limit without derivatives.
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I need to find the limit of sin(3t)^2/t^2 as t approaches 0. We have not yet learned L'Hopital's rule, so how do I find the limit here?

I tried to take the derivative of sin(3t)^2/t^2, but it is nowhere near cancelling out 't' from the bottom.

Thanks
 
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Do you know the limit of sin(x)/x as x->0? Do you know the rules for a limit of a product?

If, so use those along with the fact that \frac{\sin^2(3t)}{t^2}=\frac{\sin(3t)}{t}\cdot\frac{\sin(3t)}{t}
 
And, if that isn't enough, the obvious
\frac{sin^2(3t)}{t^2}= 9\frac{sin(3t)}{3t}\cdot\frac{sin(3t)}{3t}
 
\lim_{t \rightarrow 0}(\frac{sin(3t)^2}{t^2})

Now remember \lim_{t \rightarrow 0}\frac{sin(t)}{t}=1.

So \lim_{t \rightarrow 0}(\frac{sin(3t)^2}{t^2})=\lim_{t \rightarrow 0}(\frac{sin(3t)}{t})^2.

It is pretty easy from now on. In future, please consider using LaTeX code.

Regards.
 
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