# How do you find the limit of #tan(4alpha)/sin(2alpha)# as #alpha->0#?

We do it algebraically, using:

We can say:

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To solve without using derivatives (l'Hospital's Rule) see below.

Use the sine of a double angle formula to rewrite this

So

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We can rearrange the given limit to make these appear:

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To find the limit of tan(4alpha)/sin(2alpha) as alpha approaches 0, we can use L'Hôpital's rule. Taking the derivative of both the numerator and denominator with respect to alpha, we get:

d/dalpha(tan(4alpha)) = 4sec^2(4alpha) d/dalpha(sin(2alpha)) = 2cos(2alpha)

Now, substituting alpha = 0 into the derivatives, we have:

d/dalpha(tan(4alpha)) = 4sec^2(0) = 4 d/dalpha(sin(2alpha)) = 2cos(0) = 2

Therefore, the limit of tan(4alpha)/sin(2alpha) as alpha approaches 0 is equal to the limit of the derivatives, which is 4/2 = 2.

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