Find a and limit?

**fardeen_gen** · May 11th 2009, 07:53 PM

If $\lim_{\alpha\rightarrow \infty} \frac{\alpha^2 + 1}{\alpha + 1} - a\alpha$ exists, find $a\in \mathbb{R}$ and the limit.

EDIT: It has to be $\alpha\rightarrow \infty$ . Nice book I have *Dripping in sarcasm* And I agree with the finite part of course.

**NonCommAlg** · May 11th 2009, 10:35 PM

Originally Posted by fardeen_gen

If $\lim_{n\rightarrow \infty} \frac{\alpha^2 + 1}{\alpha + 1} - a\alpha$ exists, find $a\in \mathbb{R}$ and the limit.

i think you meant $\alpha \to \infty$ and not $n \to \infty.$ also by "exists" you mean "being a finite number". to me if the limit is infinity, then it still exists. if the one sided limits are not equal,

then the limit doesn't exist. other cases that i'd say the limit doesn't exist, are limits like: $\lim_{x \to\infty} \sin x$ or $\lim_{n\to\infty} (-1)^n,$ etc. enough about lecturing you! haha

to answer your question, we have $\frac{\alpha^2 +1}{\alpha + 1} - a \alpha = \frac{(1-a)\alpha^2 + 1 - a \alpha}{\alpha + 1}.$ it's clear now that the limit would be infinity unless $a=1,$ and for $a=1$ the limit is -1.

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