Square rooting of asymptotes in curve sketching (1 Viewer)

mrpotatoed · Oct 3, 2015

If you have an obligue asymptote in a graph, say y=x, and you are graphing "y=square root f(x)", does the asymptote become y=rootx? Just asking cause I'm looking as solutions where they don't change the asymptote but that seems weird

Drsoccerball · Oct 3, 2015

mrpotatoed said:
If you have an obligue asymptote in a graph, say y=x, and you are graphing "y=square root f(x)", does the asymptote become y=rootx? Just asking cause I'm looking as solutions where they don't change the asymptote but that seems weird

It should change the asymptote aswell ?

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

$x \rightarrow \infty f(x) \rightarrow x$

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

$x\rightarrow \infty \sqrt{f(x)} \rightarrow \sqrt{x}$

Squar3root · Oct 3, 2015

square root ayylmao

kawaiipotato · Oct 3, 2015

You could also think of it like this.
"If my curve's values tend to infinity but it's always hugging the y = x line, then square rooting those values will make it go under the y = sqrtx curve"
Typing that out made me realize it sounds kind of stupid but its how I remembered it lol

InteGrand · Oct 3, 2015

$I think it suffices to show that if $L_1 :=\lim _{x\to \infty}(f(x)-x)=0$, then $L_2:=\lim_{x\to \infty} \left(\sqrt{f(x)}-\sqrt{x}\right)=0$ (i.e. if the graph of $y=x$ is an asymptote of the graph of $y=f(x)$, then the graph of $y=\sqrt{x}$ is an asymptote to the graph of $y=\sqrt{f(x)}$).$

$Proof. Suppose $L_1 :=\lim _{x\to \infty}(f(x)-x)=0$. To prove the requirement, we must show that for any $\varepsilon_2 >0$, there exists $M_2>0$ such that for all $x> M_2$, we have $-\varepsilon_2 <\sqrt{f(x)}-\sqrt{x}<\varepsilon_2 $. Let $\varepsilon_1 >0$. From assumption of $L_1$ and definition of limits, we have that for some $M_1 >0$ we have for all $x>M_1$ that $x-\varepsilon_1 < f(x) < x+\varepsilon_1$.$

$Clearly by taking $M_1$ sufficiently large, all terms in this inequality are positive, so we can take square roots across it to obtain$

$\sqrt{x-\varepsilon_1} < \sqrt{f(x)} < \sqrt{x+\varepsilon_1}\Rightarrow \sqrt{x-\varepsilon_1}-\sqrt{x} < \sqrt{f(x)} -\sqrt{x}< \sqrt{x+\varepsilon_1}-\sqrt{x},x>M_1$. \ \(*)$

$Now, using the result that $\sqrt{a}-\sqrt{b}<\sqrt{a-b}$ if $a>b>0$ (proved at the end), we have that $\sqrt{x+\varepsilon_1}-\sqrt{x}<\sqrt{(x+\varepsilon_1)-x}=\sqrt{\varepsilon_1}$, and $\sqrt{x}-\sqrt{x-\varepsilon_1}<\sqrt{\varepsilon_1}\Rightarrow \sqrt{x-\varepsilon_1}-\sqrt{x} >-\sqrt{\varepsilon_1}$. Putting these in to (*), we have$

$$-\sqrt{\varepsilon_1}<\sqrt{f(x)} -\sqrt{x}<\sqrt{\varepsilon_1},x>M_1$.$

$Since given any $\varepsilon_2 >0$, we can write it in the form $\sqrt{\varepsilon_1}$ ($\varepsilon_1$ is simply $\varepsilon_2^2$), we have.$

$$-\varepsilon_2<\sqrt{f(x)} -\sqrt{x}<\varepsilon_2,x>M_1$.$

$This proves the requirement.$

$Here is a proof that $\sqrt{a}-\sqrt{b}<\sqrt{a-b}$ if $a>b>0$:$

$RHS^2 - LHS^2 = a-b - a -b + 2\sqrt{ab}$

$=-2b+ 2\sqrt{ab}$

$=2\left( \sqrt{ab}-b\right)$

$>2\left( \sqrt{bb}-b\right)$, as $a>b$

$=0$

$$\Rightarrow RHS>LHS$ as $LHS$ and $RHS$ are both positive. This completes the proof.$

Square rooting of asymptotes in curve sketching (1 Viewer)

mrpotatoed

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Drsoccerball

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Squar3root

realest nigga

kawaiipotato

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InteGrand

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