Math Limits Question

Find the limit:
lim x->5 (sin(x-5))/(x^2+3x-40)

How would i solve this problem?

lim(x->5)[sin(x - 5) / (x^2 + 3x - 40)] =

lim(x->5)[sin(x - 5) / ((x - 5)(x + 8))] =

lim(x->5)[(sin(x - 5) / (x - 5)) * (1/(x + 8))] =

lim(x->5)[sin(x - 5) / (x - 5)] * lim(x->5)[1 / (x + 8)].

Now in the first of these two limits let u = x - 5. Then since u -> 0 as x -> 5 we
can represent the first limit as a limit as u -> 0, giving us

lim(u->0)[sin(u) / u] * lim(x->5)[1 / (x + 8)] = 1 * (1/13) = 1/13.

I am assuming that you are familiar with the well-known limit lim(u->0)(sin(u)/u) = 1.
Also, the change of variable method, also called the substitution method, is a
standard method that is very useful in solving many types of math problems, from
limits to sums to calculus, to name just a few types.

Because the lim x->5 is equal to 0 for the numerator and the denominator L'Hospital's rule applies.

Take the derivative of the numerator and denominator separately:
cos(x-5)/(2x+3)

Then evaluate the limit:
lim x->5 cos(x-5)/(2x+3) = cos(0)/(10+3) = 1/13

This problem only required the use of L'Hospital's rule once but for many problems you will have to use the rule multiple times. As long as the lim of the numerator and denominator are equal, take the derivative of each, and then take the limit. Repeat as many times as necessary.