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Numerade Educator



Problem 45 Medium Difficulty

Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If l'Hospital's Rule doesn't apply, explain why.

$ \displaystyle \lim_{x\to 0} \sin 5x \csc 3x $


This limit has the form $0 \cdot \infty$. We'll change it to the form
$\frac{0}{0} \cdot \lim _{x \rightarrow 0} \sin 5 x \cos 3 x=\lim _{x \rightarrow 0} \frac{\sin 5 x}{\sin 3 x} \triangleq \lim _{x \rightarrow 0} \frac{5 \cos 5 x}{3 \cos 3 x}=\frac{5 \cdot 1}{3 \cdot 1}=\frac{5}{3}$


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Video Transcript

those problems were going to be possibly using the look tells road to solve the problem, but it's not always necessary. So we see that the function were given that we want to find the limit of Is sign five x. At times if you can three X. Which is the same thing as over the sign of three acts, then we plug in zero. We see that we would get zero or zero. This is an indeterminant form. So we can take the derivative of the numerator that would end up giving us five co sign Perfect. And then in the denominator that would give us three co sign three acts as we see. We can actually use local trials role in this case because we get the indeterminate form. Uh and then when we plug in zero, this time we get five times coastline of zero, which is just five and three times co sign of zero, which is three. So five thirds will be the final answer.