limit question?

Question

lim x -> 5- e^x/(x-5)^3. I got oo.

Answer 1

That function has an asymptote at x=5. Approaching from the negative side, the numerator is positive and the denominator is negative, so you are approaching -∞ (negative infinity).

Answer 2

well as noted x is approaching 5 from the left

so you get a finite positive # in the numerator and the denominator approaches 0 from the left as x approaches 5 from the left
so cutting corners and playing fast n loose with terminology you basically have a limit that approaches negative infinity

Answer 3

Use L'Hopital's Rule on the denominator (the derivative of the numerator is the same)

1st try 3(x-5)^2
2nd try 6(x-5)
3rd try = 6

So the limit would be e^5 / 6

Answer 4

the problem is that as x-->5 the limit approaches -oo or +oo depending on what side of 5 you are approaching from.
This means that there is no limit for x-->5

Answer 5

The top is a big number, the bottom is a negative number close to zero. The limit does not exist (it is negative infinity). [assuming you meant by x -> 5- that it approaches 5 from the left]