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I am trying to solve the following limit:

$$ \lim_{x \to 1^+} \frac{x^x+\frac{x}{2}-\sqrt x-\frac{1}{2}}{\log(x)-x+1} $$

My attempt was to substitute $x$ with $1+y$, which results in the following:

$$ \lim_{y \to 0^+}_{y<1} \frac{(y+1)^{y+1}+\frac{y}{2}-\sqrt{1+y}}{\sum_{n=1}^\infty (-1)^n\frac{y^{n+1}}{n+1}} $$

I strongly suspect that the limit does not exists (the term seems to diverge towards negative infinity), but I don't know how to proceed from here. I think that it might be easiest to find a null sequence to substitute $y$ with, that shows that the values of the term above are unbounded, but I am stuck at this for some time now.

Could somebody point me into the right direction?

UPDATE: Unfortunately I am not allowed to use De L'Hopital's rule yet.

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Are you allowed to use Taylor series? –  Dan Petersen Jan 18 '12 at 9:28
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3 Answers 3

up vote 2 down vote accepted

If you can't use L'Hopitals rule then your best bet is to use series expansion about x=1. This is basically what you're already doing with your transform $y=1+x$.

The tricky part is that you're going to have to use the binomial theorem on two of the bits on the top line:

$$(1+y)^{(1+y)} = 1 + (1+y)*y + \ldots = 1 + y + y^2 + \ldots$$

and

$$\sqrt{1+y} = (1+y)^{1/2} = 1 + (1/2)y + \ldots$$

Then your top line becomes

$$ \left(1 + \frac{y}{2} + O(y^2)\right) + \frac{y}{2} - \left( 1 + \frac{y}{2} + O(y^2) \right) = \frac{y}{2} + O(y^2) $$

The bottom, expaning your sum is

$$-y^2 + O(y^3)$$

So overall your fraction becomes:

$$\frac{\frac{y}{2} + O(y^2)}{-y^2 + O(y^3) } = \frac{-1}{2y}\frac{1 + O(y)}{1 +O(y)} = \frac{-1}{2y}(1+O(y))$$

Its clear that this will behave like $\frac{-1}{2y}$ for $y$ near 0.

Thus the result will be unbounded.

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Thank you very much! I can work with that. –  Niklas B. Jan 18 '12 at 9:32
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You should apply de L'Hospital rule. You will get

$$ \lim_{x \to 1^+} \frac{x^x+\frac{x}{2}-\sqrt x-\frac{1}{2}}{\log(x)-x+1}= \lim_{x \to 1^+} \frac{x^x(\log(x)+1)+\frac{1}{2}-\frac{1}{2\sqrt x}}{\frac{1}{x}-1}=-\infty $$

The other way is using Taylor series as follows

$$ \lim_{x \to 1^+} \frac{x^x+\frac{x}{2}-\sqrt x-\frac{1}{2}}{\log(x)-x+1}= \lim_{x \to 1^+} \frac{1+(x-1)+\frac{x}{2}-1-\frac{x-1}{2}-\frac{1}{2}}{x-1-\frac{1}{2}(x-1)^2-x+1}=-\infty $$

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Thanks for the anwers! However, please see my updated question. Unfortunately we are not allowed to use De L'Hopital yet. Also, it should be $-\infty$, because $\frac{1}{x} - 1 < 0$. –  Niklas B. Jan 18 '12 at 9:16
    
@Niklas: Just to know. Are you aware of Taylor series? I see you used one. –  Jon Jan 18 '12 at 9:26
    
Yes I am. However, for my expansion I didn't need one, we proved that as a separate theorem. –  Niklas B. Jan 18 '12 at 9:33
    
@Niklas: Ok, then we should be done. –  Jon Jan 18 '12 at 9:34
    
Thank you for this alternative solution. –  Niklas B. Jan 18 '12 at 9:35
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Use L'Hopitals rule, which basically says:

If $\lim_{x\to1} f(x) = 0$ and $\lim_{x\to1} g(x) = 0$, then

$$ \lim_{x\to1^+} \frac{f(x)}{g(x)} = \lim_{x\to1^+} \frac{f'(x)}{g'(x)} $$

You should find for your function $f'(1) \neq 0$ and $g'(1)=0$ so that the limit is unbounded.

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That'd be possible, but we haven't proven and thus are not allowed use L'Hopital yet, unfortunately. I'll edit the answer to mention this. –  Niklas B. Jan 18 '12 at 9:10
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