L'Hopital's Rule. Example Problem

L'Hopital's Rule. Example Problem

lim_(x->0) (cos(3x)-e^4x)/(x^2 -3x)
lim_(x->0) (-3sin(3x)-4e^4x)/(2x-3)=-4/-3 =4/3
Let us consider the following example of L'Hopital's rule.
If you evaluate the cosine of 0, you get 1. The exponential of 0 is 1, divided by 0 minus 0 – we get
0/0.
Factoring probably won't work, since there are no algebraic methods or anything that will help us.
We'll use L'Hopital's rule instead.

Thus, we take the derivative of the numerator (3sin3x), times 3 because of the chain rule, but you
already knew that.
The derivative of e to the 4x 1S equal to 4 times e to the 4x divided by 2 times x minus 3. In 3 other
words, we can take the limit of each factor separately, then combine them when we have a limit on
both sides.
So you plug in zero for the x-value. The sine of 0 is 0, minus e to the 0 is 1, times 4 is -4, divided
by 0 minus 3 is -3.

So the limit would be 4/3. Here it is: L'Hopital's rule.