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

1 Answer
Zack M.
Dec 14, 2014

First, lets look at a graph of the equation.

Geogebra

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.

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