Can someone help me with this calc problem?

lim as x->0 (tan^3 (2x))/(x^3)

The power series for tan y is

y + y^3/3 + 2 y^5/15 .... for small y
so the biggest term in tan 2x would be 2x for x <1
so the biggest term in tan^3 2x would be 8 x^3
and 8 x^3/x^3 = 8

Ohhh, thanks Damon!

Sure! To find the limit as x approaches 0 of (tan^3(2x))/(x^3), we can use L'Hopital's rule, which states that if both the numerator and the denominator of the fraction approach zero or infinity, then the limit can be found by taking the derivative of the numerator and denominator and then evaluating the limit again.

Let's start by taking the derivative of the numerator and denominator separately. The derivative of the numerator, tan^3(2x), can be found using the chain rule. We have:

d/dx (tan^3(2x)) = 3tan^2(2x) * sec^2(2x) * 2

Simplifying this, we get:

6tan^2(2x) * sec^2(2x)

Now, let's take the derivative of the denominator, x^3:

d/dx (x^3) = 3x^2

Now, we have the derivatives of both the numerator and denominator. We can rewrite the original expression as:

lim as x -> 0 (6tan^2(2x) * sec^2(2x))/(3x^2)

Now, we can simplify this expression by canceling out the common factor of 3:

lim as x -> 0 (2tan^2(2x) * sec^2(2x))/(x^2)

Finally, we can evaluate the limit as x approaches 0. We can substitute 0 for x in the expression and simplify further:

lim as x -> 0 (2tan^2(2x) * sec^2(2x))/(x^2)

= (2tan^2(0) * sec^2(0))/(0^2)

= (2 * 0 * 1)/(0)

= 0

Therefore, the limit as x approaches 0 of (tan^3(2x))/(x^3) is 0.