lim
x--> - infinity (x^3 + e^-x)
x--> - infinity (x^3 + e^-x)
-
Lim x-> -∞ [x^3 + 1/e^x]
If we naively sub in -∞, we get:
L = (-∞)^3 + 1/e^(-∞)
L = (-∞)^3 + e^(∞)
Now (-∞)^3 = -∞ and e^(∞) = ∞, thus the question is basically just to determine whether e^(∞) is approaching infinity faster or slower that (-∞)^3 is approaching negative infinity.
In these kinds of questions, the exponential pretty much always wins. x^3 is growing with an exponent of 3 no matter what, whereas e^(x) is growing with an exponent of increasingly large numbers. Even if the exponent on the x was 101, the exponential function would still win.
Thus, the limit is infinity.
Done!
P.S. If this was the limit as x goes to positive infinity, then the limit would still be infinity, because 1/e^x goes to 0 and x^3 goes to infinity.
If we naively sub in -∞, we get:
L = (-∞)^3 + 1/e^(-∞)
L = (-∞)^3 + e^(∞)
Now (-∞)^3 = -∞ and e^(∞) = ∞, thus the question is basically just to determine whether e^(∞) is approaching infinity faster or slower that (-∞)^3 is approaching negative infinity.
In these kinds of questions, the exponential pretty much always wins. x^3 is growing with an exponent of 3 no matter what, whereas e^(x) is growing with an exponent of increasingly large numbers. Even if the exponent on the x was 101, the exponential function would still win.
Thus, the limit is infinity.
Done!
P.S. If this was the limit as x goes to positive infinity, then the limit would still be infinity, because 1/e^x goes to 0 and x^3 goes to infinity.
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This is infinity (because the exponential grows faster than any polynomial).
I hope this helps!
I hope this helps!