# Evaluation of limits

I'm having trouble identifying some limits, or actually, I figured out what they are just not the argument for why. Here they go:

$$\lim_{a\to1}\frac{1}{a}\left(x\cdot\left(1-a\right)^{\frac{1-a}{a}}+b\left(1-a\right)^{\frac{1}{a}}\right)^{a}=x$$

And the limit: $$\lim_{a\to-\infty}\frac{1-a}{a}\left(\frac{bx}{1-a}+1\right)^{a}=-e^{-bx}$$

So the first is easy if one may evaluate the terms in parts. I know that if $a_{n},b_{n}$ are converges to a,b then $$a_{n}b_{n}\to ab\quad a_{n}+b_{n}\to a+b$$ but i'm not sure how to handle the power. The second one I feel like doing the good old $e^{\ln(x)}=x$ trick which gives: $$e^{a\cdot\left(\left(\ln\left(\frac{bx}{1-a}\right)+1\right)+\ln\left(\frac{1-a}{a}\right)\right)}$$

But it doesn't really work out, the second terms looks delicious since it's limit on its self is $\pi \iota$ but the "a" in front kind of messes that up and the first part looks like something that could easily be l'hospital'ed but I just can't get the last steps.

Hope someone is willing to help.

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I think you can see the expression in the first limit as $(b$+\frac {x}{1-a}$$)^{a}\frac {1-a}{a}. Check it again,may be this latter one can be aplicable for you. – Babak S. May 4 '12 at 18:31 I find it sadistic to give such limits for a student (at least, the first one is just mean). – Pedro Tamaroff May 4 '12 at 18:36 haha i agree with you! But it is though in an applied course - so that's kind of an excuse. – Henrik May 4 '12 at 18:39 @Sorouh Actually your suggestion was the assignment, i rewrote it to what I have (feeling that it was easier). Can you find the limit directly from your expression? – Henrik May 4 '12 at 18:41 Take this one for the second: ((\frac {bx}{1-a}+1)^{1-a})^{-1}.(\frac {bx}{1-a}+1).(\frac {1-a}{a}). the first part would be e^{-bx}. :) – Babak S. May 4 '12 at 19:11 ## 1 Answer$$\lim_{a\to1}\frac{1}{a}\left(x\cdot\left(1-a\right)^{\frac{1-a}{a}}+b\left(1-a\right)^{\frac{1}{a}}\right)^{a}=x $$What happens when you put a=1 in your expression? You get:$$\frac{1}{1}\left(x\cdot\left(1-1\right)^{\frac{1-1}{1}}+b\left(1-1\right)^{\frac{1}{1}}\right)^{1}=x \frac{1}{1}\left(x\cdot0^{0}+b\cdot 0^{1}\right)^{1}=x $$You have a 0^0 indetermination. So let's tweak the expression. First, let's change a-1=t, to get$$\mathop {\lim }\limits_{t \to 0} \frac{1}{{t + 1}}{\left( {x\cdot{{\left( { - t} \right)}^{\frac{{ - t}}{{t + 1}}}} + b{{\left( { - t} \right)}^{\frac{1}{{t + 1}}}}} \right)^{t + 1}} = x$$Now we have a x^x type of function. If you change t=-u; you get$$\lim_{u \to 0} u^u=1$$(this is a known limit, I guess). Since all of the others are continuous, you get$$\mathop {\lim }\limits_{t \to 0} \frac{1}{{t + 1}}{\left( {x\cdot{{\left( { - t} \right)}^{\frac{{ - t}}{{t + 1}}}} + b{{\left( { - t} \right)}^{\frac{1}{{t + 1}}}}} \right)^{t + 1}} = \frac{1}{1}{\left( {x\cdot{1^1} + b \cdot {0^1}} \right)^1} = x$$Here you let 1-a=u.$$\eqalign{ & \mathop {\lim }\limits_{a \to - \infty } \frac{{1 - a}}{a}{\left( {1 + \frac{{bx}}{{1 - a}}} \right)^a} = \mathop {\lim }\limits_{u \to \infty } \frac{u}{{1 - u}}{\left( {1 + \frac{{bx}}{u}} \right)^{1 - u}} = \cr & \mathop {\lim }\limits_{u \to \infty } \frac{u}{{1 - u}}\frac{{\left( {1 + \frac{{bx}}{u}} \right)}}{{{{\left( {1 + \frac{{bx}}{u}} \right)}^u}}} = \left( { - 1} \right)\frac{{\left( {1 + 0} \right)}}{{{e^{bx}}}} = - {e^{ - bx}} \cr}\$

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Thank you man, that was perfect! – Henrik May 4 '12 at 21:02