Can someone help me with some calculus? (limits. . . I'm lost)?

Question

I'm so confused on doing these limits :/ can someone please explain. sorry there are quite a few of them :(

b cannot be 0 , evaluate lim x -> b (x^3-b^3 )/ (x^6 - b ^6)

lim x -> inifinty (e^x)/(1-x^3)

lim x-> 3+ sq root of (t^2 -9)/ (t-3)

find the limit lim x-> 5+ (5+ IxI)/( 5-x) when the IxI is the greatest integer of x

email: [e-mail]

Answer 1

Before I answer, word of advice: REMOVE YOUR EMAIL ADDRESS, or at least put spaces in it. You're going to get a ton of spam.

Now, on to your questions:

b cannot be 0 , evaluate lim x -> b (x^3 - b^3 )/ (x^6 - b ^6)

Are you sure b cannot = 0? It should also say x cannot = b.

Let's factor the original equation:

(x^3 - b^3)/(x^3 - b^3)(x^3 + b^3) = 1/(x^3 + b^3)

Now, let x = b:

1/(b^3 + b^3) = 1/2b^3

Therefore,

lim x-->b (x^3 - b^3)/(x^6 - b^6) = 1/(2b^3)

lim x -> inifinty (e^x)/(1 - x^3) = 0

Sorry, can't explain, I had to take the derivative and do some manipulation.

lim x-> 3+ sq root of (t^2 - 9)/(t - 3)

lim x-->3+ (from the right) sqrt[(t^2 - 9)/(t - 3)]

Factor:

lim x-->3+ sqrt[(t - 3)(t + 3)/(t - 3)]

lim x-->3+ sqrt(t + 3) = sqrt(3 + 3) = sqrt6

find the limit lim x-> 5 + (5 + IxI)/(5 - x) when the IxI is the greatest integer of x

As x--> 5, the numerator approaches 10 and the denominator approaches 0. 10 divided by a number really really small (approaching 0) = infinity.

Answer 2

i didn't feel like reading it all, but u probably need l'hopital's rule