What is the limit as x approaches infinity of #x^(ln2)/(1+ln x)#?

Answer 1
#lim_{x \to +oo} \frac{x^{ln 2}}{1+ln x} = +oo#

This can be found by using L'Hospital's Rule, which states that every limit of a fraction is equal to the limit of the derivatives of the fraction. More formally:

#lim \frac{f(x)}{g(x)} = lim \frac{f'(x)}{g'(x)}#
Using this rule on this problem: #lim_{x \to +oo} \frac{x^{ln 2}}{1+ln x} = # # lim_{x \to +oo} \frac{ln (2)*x^{(ln 2)-1}}{\frac{1}{x}} = lim_{x \to +oo} ln(2)*x^{ln(2)-1}*x# # lim_{x \to +oo} ln(2)*x^{(ln 2)-1+1}=lim_{x \to +oo} ln(2)*x^{(ln 2)}# Now all you have to know is that when #x# goes to infinity, #x^{ln2}# also goes to infinity.
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Answer 2

The limit as x approaches infinity of x^(ln2)/(1+ln x) is infinity.

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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.

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.

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.

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.

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