in the denominator, and you do at least in number 1 and 3.So for number one we need to multiply by the conjugate to get rid of that pesky square root.[(√(x+1) - 2) / (x-3)]*[(√(x+1) + 2)*(√(x+1) + 2)]Then foil that crap out.We get:(x+1 -4) / (x-3)(√(x+1) + 2)x-3 / (x-3)(√(x+1) + 2)Cancel an x-3So we get:1/ (√(x+1) + 2)If we plug in 3, we get 1/(2+2) =1/4Number 2,......
3. Multiply and divide by (t - 5) to get rid of the complex fraction. lim{t→0} 5/[t(t - 5)(t + 1)] This limit DNE (does not exist), it approaches + and - infinity on either side.
I dont know know lhopital's rule.
But what you need to do is check if you get a 0, in the denominator, and you do at least in number 1 and 3.
So for number one we need to multiply by the conjugate to get rid of that pesky square root.
[(√(x+1) - 2) / (x-3)] * [(√(x+1) + 2)*(√(x+1) + 2)]
Then foil that crap out.
We get:
(x+1 -4) / (x-3)(√(x+1) + 2)
x-3 / (x-3)(√(x+1) + 2)
Cancel an x-3
So we get:
1/ (√(x+1) + 2)
If we plug in 3, we get 1/(2+2) = 1/4
Number 2, we can just plug in 2 since the denominator deos not go to 0
1/2+ 1/2 / (2+2)
1/4
Number 3, we get that the bottom goes to 0, but the top does not... This means we have a limit that is either does not exist, or is -∞ or ∞
We check by doing a sign check.
5 / -1 / 0
Unfortunately theres no way to check because we would get -∞ from one side and ∞ from the other. Therefore the limit as t--> 0 would be does not exist!
Do you know l'Hopital's Rule? It applies to all of these.
1. lim (x->3) (sqrt (x + 1) - 2) /(x - 3) = lim (x->3) 1/(2sqrt(x + 1))/1
=
1/(2sqrt(3 + 1))/1 = 1/(2*2) = 1/4
The others follow in a similar way
