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How do you find the limit of #(x^(3 sin x))# as x approaches 0+?

Answer 1

Elijah Abram

#1#

#x^(3 sin x)equiv (1+delta)^(3sin(1+delta))#

# (1+delta)^(3sin(1+delta)) = 1 + 3sin(1+delta) delta + (3sin(1+delta)(3sin(1+delta)-1))/(2!)delta^2+cdots#

#lim_(x->0)x^(3 sin x) equiv lim_(delta->-1) (1+delta)^(3sin(1+delta)) =1#

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Answer 2

Ezekiel Alexander

To find the limit of (x^(3 sin x)) as x approaches 0+, we can use the squeeze theorem.

First, we need to establish the bounds for the function. Since -1 ≤ sin x ≤ 1 for all x, we have -x^3 ≤ x^(3 sin x) ≤ x^3.

Now, as x approaches 0+, both -x^3 and x^3 approach 0. Therefore, by the squeeze theorem, the limit of x^(3 sin x) as x approaches 0+ is also 0.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.