Understanding Limits: Evaluating lim x->∞ and lim x->-27 for 2 Questions

[kiss89]

I have 2 questions about limits :

1)Evaluate : lim as x approaches infinity for
(3x-4-4x^2) / (x^2 -16)

2)Evaluate: lim as x approaches -27 for
(27+x) / ( 7/2 + 3 )

thank u.

 

[JasonRox]

I have 2 questions about limits :

1)Evaluate : lim as x approaches infinity for
(3x-4-4x^2) / (x^2 -16)

2)Evaluate: lim as x approaches -27 for
(27+x) / ( 7/2 + 3 )

thank u.

If you show us some work, we will be glad to help you.

 

[kiss89]

Sorry about that...heres my work
http://img88.imageshack.us/img88/9971/img002sy9.png

 

[bob1182006]

I have 2 questions about limits :

1)Evaluate : lim as x approaches infinity for
(3x-4-4x^2) / (x^2 -16)

2)Evaluate: lim as x approaches -27 for
(27+x) / ( 7/2 + 3 )

thank u.

for #1, in order to find the limit as x approaches infinity you need to divide the numerator/denominator by the highest power of x in the denominator, in this problem it would be x^2.
Doing this you'd end up with some 1/x's or x^n times some constant (1,2,3,...) which =0 when you take their limit as x approaches infinity

for #2, lim as x approaches a for f(x) = f(a) IF a is in the domain of f

 

[turdferguson]

For #2, when you plug -27 in you get a number over a nonzero. That means the limit equals the value of the function, in this case, 0

 

[kiss89]

what about Q.1 i still don't understand how to solve it using infinity?

 

[turdferguson]

In the limit, only the highest degree matters. Because the numerator and denominator have the same degree, the function will have a limit (horizontal asymtote) approaching infinity. Divide the coefficient of the numerator by the coefficient of the denominator and you have all that's left when x is big. Your post is different from your work, but either way the answer should be apparent in 2 seconds