Ayudame!

Evaluate the following limits:

1. limit as t approaches 9: ( 9 - t) / 3 - rad t (sqr rt of t)

AND

2. limit as x > -4 (the # -4, not from the left side):

((1/4) + (1/x)) / (4 + x)



.....

it seems to me that neither limits exist.
Algebraically (unless theres some trick I missed)
and on my calculator.

when I graph and attempt to seek the value for entering x (or t for # 2)
The first one gives me y = *blank*
the second one gives me an error message

so I'm assuming I'm right in that neither limit exists.
Anyone care to dispute that?

Using L'Hopital's rule on 1. will give you 6, I think.

Edit: Graphing it confirms that it is indeed 6. Try zooming in on the neighborhood of x = 9 and you'll see the y values approaching 6.

For the 2nd limit

lim(x -> -4) ((1/4) + (1/x))/(4+x)

After fracking up my algebra the first time, I simplified it to lim(x -> -4) (x[sup]2[/sup] + 8x + 16)/4x. How do you evaluate these limits again?

For the 1st limit I get... 1/(1/6) which is...6

Dear me that's something I haven't heard in a good few years. Good ol limit state calculus badness.

And 2. comes out to be -(1/16) just by algebraic simplification.

I don't think I'vel earned L'Hopital's rule yet...

And I think I need a run through the algebra on the second one.
I'm trying to work it out now, maybe I suck.
[edit]
nevermind, I completely forgot that a / b + c / d = (ad + cb) / bd, but I found it online.
...

I applied that rule once.

then I applied it to the top only, and i got this:
(4x+x^2)*(16+4x)/(4x)) / (4+x)^2

=

(x(4 + x) * (16 + 4x) / 4x) * (1 / ((4+x)^2))

=

(x(16 + 4x) / (4x(4 + x)) = (4x(4+x)) / (4x(4+x) = 1
*scratches head*
I guess I gotta go back over it again.

L'Hoptial's rule = when a limit is in indeterminate form, you can take the limit of the derivative of the numerator divided by the derivative of the denominator to find the limit of the original.

Wait, is the first one supposed to be (9 - t) / (3 - sqrt t) ?

Without L'Hospital's, the first can be simplified once you rewrite the numerator as (3^2 - sqr(t)^2). That can be factored in (3 - sqr(t))(3 + sqr(t)). Then (3 - sqr(t)) cancels out with the denominator and you're left with (3 + sqr(t))/1, so plug in 9 and you've got 6.

The second one is solvable with the ad + bc/bd rule, as has been said. ((1/4) + (1/x)) becomes (4 + x)/(4x), which is then divided by (4 + x), canceling out the numerator and leaving you with 1/4x.