How do I solve this polynomial limit?

[greg_rack]

Homework Statement

Same as the other thread I opened... this function is literally messing me up :)
$$\lim_{x \to +\infty }\sqrt{x^{2}-2x}-x+1$$

Relevant Equations

none

I'll write my considerations which lead me to get stuck on the $\infty-\infty$ form.
$$\lim_{x \to +\infty }\sqrt{x^{2}-2x}-x+1 \rightarrow |x|\sqrt{1-0}-x+1$$
And I have no idea on how to go on...

 

[fresh_42]

You could rewrite it:
\begin{align*}
\lim_{x \to \infty}\left(\sqrt{x^2-2x}-x+1\right)&=\lim_{x \to \infty}\left(\sqrt{(x-1)^2-1}-(x-1)\right)\\&=\lim_{y \to \infty}\left(\sqrt{y^2-1}-y\right)\\
&=\lim_{y \to \infty}\left(\sqrt{y^2-1}-\sqrt{y^2}\right)
\end{align*}
and consider whether the difference between these two roots tend to zero or not.

 

[anuttarasammyak]

$$\sqrt{x^2-2x}-x+1=\frac{x^2-2x-(x-1)^2}{\sqrt{x^2-2x}+x-1}$$

 

[Delta2]

You almost did all the job.. Because we are taking the limit at $x\to+\infty$ , x is positive in a neighborhood of $+\infty$ thereforce you can replace $|x|=x$.

 

[greg_rack]

You almost did all the job.. Because we are taking the limit at $x\to+\infty$ , x is positive in a neighborhood of $+\infty$ thereforce you can replace $|x|=x$.

So we obtain that limit tending to 1, right? $x-x+1$ indeed... but it doesn't work and I can't get what's wrong

 

[greg_rack]

and consider whether the difference between these two roots tend to zero or not.

Oh yes, it does tend to zero now, and that should be correct... but why was the $x-x+1$ wrong? All I did was simply take the $x^2$ out of the root

And what about the limit with $x \to -\infty$ of that same function? I'm stuck again, since the form now is $\infty+\infty$

 

[fresh_42]

Oh yes, it does tend to zero now, and that should be correct... but why was the $x-x+1$ wrong? All I did was simply take the $x^2$ out of the root

And what about the limit with $x \to -\infty$ of that same function? I'm stuck again, since the form now is $\infty+\infty$

You did more. You had $x\sqrt{1-\dfrac{2}{x}}-x+1$ and only let the $2/x$ term go to infinity, so you got indeed $\infty-\infty+1$ which could be any number.

 

[greg_rack]

You did more. You had $x\sqrt{1-\dfrac{2}{x}}-x+1$ and only let the $2/x$ term go to infinity, so you got indeed $\infty-\infty+1$ which could be any number.

Thanks man, now I got what I was doing wrong!

 

[kuruman]

From post #2 by @fresh_42,
$$\sqrt{y^2-1}-\sqrt{y^2}=\frac{\left(\sqrt{y^2-1}\right)^2-\left(\sqrt{y^2}\right)^2}{\sqrt{y^2-1}+\sqrt{y^2}}.$$Simplify the numerator and consider what happens to the denominator in the limit $y \rightarrow \infty.$

 

[fresh_42]

And what about the limit with $x \to -\infty$ of that same function? I'm stuck again, since the form now is $\infty+\infty$

This case is easier. One blow up plus a second blow up is still an explosion. You only have to pay attention if terms seem to cancel each other. In such a case you have to look closer on the degree or speed of cancellation. $\infty - \infty$ and $\dfrac{\infty}{\infty}$ are not defined, i.e. you have to transform the expression until they are either defined or you can otherwise come to a conclusion. E.g. if you take my transformation with the $y$ and combine it with the trick in post #3: $\sqrt{...}-\sqrt{...}=\dfrac{\sqrt{...}^2-\sqrt{...}^2}{\sqrt{...}+\sqrt{...}}$ then you get a form where it is clear how it behaves.