Help With Evaluating Limits Involving Infinity

[Laucchi]

I'm doing a review of Limits Involving Infinity, and I'm a bit unsure with this one.
limit as x approaches infinity of x³/e^x​
for this one, I'm a bit confused, because both of these are dominant terms. How would I go about solving it?
I've already graphed it:

Thank You!

 

[srmichael]

I'm doing a review of Limits Involving Infinity, and I'm a bit unsure with this one.
limit as x approaches infinity of x³/e^x​
for this one, I'm a bit confused, because both of these are dominant terms. How would I go about solving it?
I've already graphed it:
View attachment 2596
Thank You!

Well, what does it look like in the graph that the function is approaching as x goes to infinity?

 

[Laucchi]

Would it be zero (0)? Because on the right side x is unbounded and the y-values are getting closer and closer to zero?

 

[srmichael]

Would it be zero (0)? Because on the right side x is unbounded and the y-values are getting closer and closer to zero?

Yes. In this case, you can think of it that as x gets very large, e^x will get larger faster than x^3. And as we know the larger the denominator becomes the more it approaches 0.

 

[Laucchi]

Thank you very much!

 

[JeffM]

I'm doing a review of Limits Involving Infinity, and I'm a bit unsure with this one.
limit as x approaches infinity of x³/e^x​
for this one, I'm a bit confused, because both of these are dominant terms. How would I go about solving it?
I've already graphed it:
View attachment 2596
Thank You!

Another way: use L'Hospital's Rule repeatedly

\(\displaystyle \displaystyle \lim_{x \rightarrow \infty}\dfrac{x^3}{e^x} = \lim_{x \rightarrow \infty}\dfrac{3x^2}{e^x} = \lim_{x \rightarrow \infty}\dfrac{6x}{e^x} = \lim_{x \rightarrow \infty}\dfrac{6}{e^x} = 0\)