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.

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.

anonymous · Answer

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

anonymous · Answer

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).

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

so you're saying it will look like this:

$$\frac{ x+3 }{x+3 } $$

and then plug in "-3" for x?