Question

how would I take the limit of Tan(4x)/Sin(3x) as x->o

Answer 1

Do you know L’Hospital’s Rule?

http://tutorial.math.lamar.edu/Classes/CalcI/LHospitalsRule.aspx

Answer 2

I personally do, however the class is not at that point and she won't let me use it, so using basic concepts can it be done?

Answer 3

How does your class do these problems? Perhaps, are you expected to use trigonometric identities to expand tan(4x) and sin(3x) ?

Answer 4

It's the first week of Calc 1. We learned how to do basic limits using the following methods. Pluging-in, simplifying, and squeeze theorem... I can't seem to use any of those methods to solve this... She teaches by example and nothing we did in class had Tangent in the numerator and Sine in the denominator so I'm just sitting here with an answer from L'hospital and no idea how to explain it.

Answer 5

Most likely you have to use the fact that:\[\lim_{x\rightarrow 0}\frac{\sin x}{x}=1\]

Answer 6

I ended up using L'hospital and I'm just going to stick with it.

Answer 7

using L'hospitals rule I get the limit as x approaches 0 =1 but I'm not sure how to do it otherwise

Answer 8

@jonnymiller

Using the same rule, I got 4/3.

Answer 9

Assuming what I posted earlier, you get a solution writing the equation like this:\[\frac{\tan (4x)}{\sin (3x)}=\frac{1}{\cos{4x}}\cdot 4\cdot \frac{\sin (4x)}{4x}\cdot \frac{1}{3}\cdot \frac{3x}{\sin (3x)}\] Then you take the limit as x goes to zero of each piece. The 1/cos goes to 1, the sin(4x)/4x goes to 1, and the 3x/sin(3x) goes to 1.

Answer 10

How did you get 4/3?