Question

What is the limit as x approaches 0 of `sin(3x)/(5x^3-4x)`?

Answer 1

First, lets look at a graph of the equation.

We can see that our answer should be slightly greater than `-1`.

Using L'Hospital's rule, which tells us that if the limit of two functions, f(x) and g(x) are both `0` or `+- oo`, then;

`lim (f(x))/(g(x)) = lim (f'(x))/(g'(x))`

Let `f(x) = sin(3x)` and `g(x) = 5x^3 - 4x`. Both are equal to `0` when `x = 0`, therefore, L'Hospital's rule applies and we can take the derivative of each. Using the chain rule;

`f'(x) = 3 cos(3x)`

and using the power rule;

`g'(x) = 15x^2 - 4`

Now we can take the limit of our new function using L'Hospital's rule.

`lim_(x->0) (f'(x))/(g'(x)) = (3cos(3x))/(15x^2 - 4) = -3/4`

This appears to agree with our graphical approximation.