Question

Limits, Cal 1
lim √(x+19) -4 / (x²+3x)
x-->3

My teacher wants us to TRY these but she is out of town so she cannot answer our questions, our library is under construction and that is where the tutoring takes place so they are not available and my book has not come in. No one can seem to help me with this problem. Can anyone show a step by step solution? I am stumped :/ Idk how to start at all.

Answer 1

And that is supposed to be "x--> -3" not "x-->3"

Answer 2

Using L'Hopital's rule, lim(x-->-3) of f(x)/g(x) = lim(x-->-3) of ((d/dx)f(x)/(d/dx)g(x)).

When I do this I get 1/[2*((x+19)^(1/2))*(2x+3)]. Taking the limit of this as x goes to -3, I get -(1/24).

Answer 3

do you know l'hopital, or are you supposed to use something else?

Answer 4

your best bet is to factor the denominator, then multiply top and bottom by \(\sqrt{x+19}+4\)

Answer 5

then numerator will be \(x+3\) which will cancel with the factor of \(x+3\) in the denominator
then you can plug in the \(-3\) to get your answer

Answer 6

so you're saying it will look like this:

\[\frac{ x+3 }{x+3 } \]

and then plug in "-3" for x?