Question

I need some help on this question!
lim(x-> -Infinity) 7/e^x + 6

Answer 1

What does e^x approach when x ===> infinity?

Answer 2

When infinity is positive e^x approaches 0

Answer 3

What's e to the power of a large number? It's not zero

Answer 4

Oh okay, the then e to a large number is a larger number

Answer 5

And 7 divided by some huge number is close to...?

Answer 6

hint : use log functions to solve this say y= 7/e^x+6 and then ln y = ln .....and evaluate the limits., other wise you can get an answer in one step , provided you know L'Hospitals rule

Answer 7

You could use logs, but don't need to, it's easy enough analytically.

Answer 8

Not sure if logs would be helpful... ln(7/e^x+6) isn't separable. Just evaluate the limits separately... limit of 7/e^x + limit of 6

Answer 9

logs would be INDEED helpful, provided you know what you are doing !

Answer 10

but I will let you carry on with your method :D

Answer 11

You don't need logs. Evaluate them separately, analytically, just using basic limit rules.

Answer 12

You don't need l'hopital's rule either.

Answer 13

lim(x-> -Infinity) 7/e^x + 6

if e^x approaches infinity, then 7/(infinity) = ...? and then you have your answer (what's leftover?)

Answer 14

anything divided by infinity is zero right?

Answer 15

Yes, so what's left?

Answer 16

lim(x-> -Infinity) 0 + 6

Answer 17

7/6 would be the answer correct?

Answer 18

Oh i just realized it's negative infinity in the original problem....? then 7/e^x would approach infinity.

If it was positive infinity like i thought, the limit would be 6.