please help! I don't know which method to use. do I use the ratio test, the integral test, or what?
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lim x-->∞ (x^n)/(e^x) = 0
In this case, we wouldn't use the ratio test or the integral test because both of those tests are for converging/diverging series.
Just based on inspection, the denominator (e^x) will increase faster than the numerator (x^n), no matter what n is. As x goes to infinity, the denominator will get larger faster than the numerator will, so the function will eventually go to 0. This is not a formal proof though.
We cannot plug in x =∞ because we would get the indeterminate form: ∞/∞.
Therefore, we can use L'Hospital's rule, which states that we can differentiate the numerator and differentiate the denominator seperately, and evaluate the limit again. What will happen here is that if you keep differentiating the numerator (x^n), you'll eventually get a constant for the numerator after differentiating n times, but the denominator (e^x) will always stay the same since d/dx(e^x) = e^x.
Try it with n = 2, for example:
n = 2 would give us x² --> 2x --> 2, and once the numerator = 2, we could evaluate lim x-->∞ 2/(e^x) because that would just = 0.
In this case, we wouldn't use the ratio test or the integral test because both of those tests are for converging/diverging series.
Just based on inspection, the denominator (e^x) will increase faster than the numerator (x^n), no matter what n is. As x goes to infinity, the denominator will get larger faster than the numerator will, so the function will eventually go to 0. This is not a formal proof though.
We cannot plug in x =∞ because we would get the indeterminate form: ∞/∞.
Therefore, we can use L'Hospital's rule, which states that we can differentiate the numerator and differentiate the denominator seperately, and evaluate the limit again. What will happen here is that if you keep differentiating the numerator (x^n), you'll eventually get a constant for the numerator after differentiating n times, but the denominator (e^x) will always stay the same since d/dx(e^x) = e^x.
Try it with n = 2, for example:
n = 2 would give us x² --> 2x --> 2, and once the numerator = 2, we could evaluate lim x-->∞ 2/(e^x) because that would just = 0.
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Use repeated iterations of L'Hopital's rule:
lim(x->inf)(x^n / e^x) = lim(x->inf)(n*x^(n-1) / e^x) = lim(x->inf)(n*(n-1)*x^(n-2) / e^x) = ...
You keep on taking derivatives for the x^n term until you are left with a constant on the top and either e^x or (x^r)*e^x on the bottom (0
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lim(x->inf)(x^n / e^x) = lim(x->inf)(n*x^(n-1) / e^x) = lim(x->inf)(n*(n-1)*x^(n-2) / e^x) = ...
You keep on taking derivatives for the x^n term until you are left with a constant on the top and either e^x or (x^r)*e^x on the bottom (0
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Chuck Norris
1
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