# Math Help - help would be very much appreciated please

1. ## help would be very much appreciated please

Evaluate the limit and indicate appropriate limit laws.

lim square root of u^4+3u+6
u->-2

Evaluate the limit

lim x^2+5x+4/x^2-3x-4
x->-4

Find the limit

lim t^2+2/t^3+t^2-1
t->-infinity

Please can someone help me?

2. Originally Posted by bballj228
Evaluate the limit and indicate appropriate limit laws.

lim square root of u^4+3u+6
u->-2

Evaluate the limit

lim x^2+5x+4/x^2-3x-4
x->-4

Find the limit

lim t^2+2/t^3+t^2-1
t->-infinity

Please can someone help me?
For the first one all you need to know is this
Let $f(x)$ be a function defined at some point a, then

$\lim_{x\to{a}}f(x)=f(a)$

For the second one, consider two polynomials

$f(x)$ and $g(x)$ if

$f(a)=g(a)=0$

Both $f(x)$ and $g(x)$ contain the factor $x-a$

Now for your last one Consider multiplying by $\frac{\frac{1}{x^2}}{\frac{1}{x^2}}$

from there you can apply the first rule I gave you remembering the lax expression $\frac{a}{\infty}=0$ where a is a nondescript constant

3. ## Check this out

Hi!bballj228
here is the solution
1))A=Lt [u^4+3u+6]^1/2
u->-2
put u =-2 and u will get
A=[16]^(1/2)
if u roughly look at the xpression or its graph u will find that the xpression>0 for all values of u.
therefor A=[16]^(1/2)=+4.
2)) B=Lt (x^2+5x+4)/(x^2-3x-4) x->-4
Nox just put x=-4 u will get
B=0/24=0
3))C=Lt[t^2+2]/t^3+t^2-1
t->infinity
divide numerator and denominator by t^2 u will get
C=Lt[1+2/(t^2)]/[t+1-1/(t^2)]
t->infinity
now put t=infinity
[remember 1/infinity=0]
finally u will get
C=1/(1+infinity) = 0
(infinity+1=infinity itself)
hope this helps ya.