Take the limit as x goes to infinity of (e^x)/(x^2)
This is an infinity/infinity problem right? If we take the derivative twice using l'hopital's rule, we end up with e^x/0... Is the answer something like going to +/- infinity?
Is it crazy to approach bringing the x of the e^x down before applying the rule?
Thanks!!
This is an infinity/infinity problem right? If we take the derivative twice using l'hopital's rule, we end up with e^x/0... Is the answer something like going to +/- infinity?
Is it crazy to approach bringing the x of the e^x down before applying the rule?
Thanks!!
-
lim x---> ∞: (e^x) / x²
You're right, it does give us ∞/∞, so we can use L'Hopital....
lim x---> ∞ (LH): (e^x) / 2x
Plugging in ∞ for all x values will give us ∞/∞ again. Using L'Hopital again....
lim x---> ∞ (LH): (e^x) / 2.
Plug in ∞ again will just give us ∞ / 2. We actually can't use L'Hoptial again because doing so will give us (e^x) / 0, which we know give us the divide by zero error. So, in this case, the answer will just be ∞.
Final Answer:
∞
You're right, it does give us ∞/∞, so we can use L'Hopital....
lim x---> ∞ (LH): (e^x) / 2x
Plugging in ∞ for all x values will give us ∞/∞ again. Using L'Hopital again....
lim x---> ∞ (LH): (e^x) / 2.
Plug in ∞ again will just give us ∞ / 2. We actually can't use L'Hoptial again because doing so will give us (e^x) / 0, which we know give us the divide by zero error. So, in this case, the answer will just be ∞.
Final Answer:
∞
-
Little Miss Sunshine -
Yes, you are definitely on the right track. However, stop taking derivatives when you get to e^x / 2 because you can't have zero in the denominator, agree?
As x goes to +infinity the limit goes to + infinity
Hope that helps
Yes, you are definitely on the right track. However, stop taking derivatives when you get to e^x / 2 because you can't have zero in the denominator, agree?
As x goes to +infinity the limit goes to + infinity
Hope that helps
-
No, I think the second derivative is e^x/2. So the limit tends to +infinity.