What is the curved asymptote of $\frac{x+1}{\sqrt{3x-2}}$?

Question

Please excuse any non-technical language I use...

I'm confused by the process for finding the curved asymptote (as $x$ gets really large) of the graph $y=\frac{x+1}{\sqrt{3x-2}}$.

I first looked at the (naive?) approach of separating the fraction up into two parts:

\begin{align} y &= \frac{x+1}{\sqrt{3x-2}} \\ &= \frac{x}{\sqrt{3x-2}} + \frac{1}{\sqrt{3x-2}} \\ \text{as } x \rightarrow \infty, \quad y &\rightarrow \frac{x}{\sqrt{3x-2}} \end{align}

Seemed ok, rationale being that the 1/sqrt part goes to zero compared to the ~ sqrt() part.

Then after a bit of searching I came to this video presented here: https://www.youtube.com/watch?v=8aWFSNbl2Ho and tried a similar method as presented...

\begin{align} y &= \frac{x+1}{\sqrt{3x-2}} \\ &= \frac{x+1}{\sqrt{3x-2}} \times \frac{\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{x}}} \\ &= \frac{\sqrt{x}+\frac{1}{\sqrt{x}}}{\sqrt{3-\frac{2}{x}}} \\ \\ \text{as } x \rightarrow \infty, \quad y &\rightarrow \sqrt{\frac{x}{3}} \end{align}

Which looks close(ish).

Wolfram Alpha however has different ideas: http://www.wolframalpha.com/input/?i=y%3D(x%2B1)%2Fsqrt(3x-2)+asymptotes

It gives the relation $x=3y^2-\frac{8}{3}$ which rearranges to a similar curve to my attempt: $$y=\frac{1}{3}\sqrt{3x-8}$$

Am I on the right track with any of my attempts? I feel like I'm closer with the 2nd attempt but I'm not allowed to disappear both the $1/\sqrt{2}$ and the $2/x$ parts like that.

Answer 1

Using equivalents, when $x$ is large $$x+1\sim x \qquad , \qquad \sqrt{3x-2}\sim \sqrt{3x}\qquad \implies y \sim \frac{x}{\sqrt{3x}}=\frac{\sqrt x}{\sqrt{3}}$$

Letting $x=\frac 1t$, you also could write $$y=\frac{1+\frac 1t}{\sqrt{\frac 3t -2}}=\frac 1{\sqrt t}\frac{1+t}{\sqrt{ 3 -2t}}$$ Now, use Taylor expansion around $t=0$ (or the generalized binomial theorem) $$\sqrt{ 3 -2t}=\sqrt{3}-\frac{t}{\sqrt{3}}-\frac{t^2}{6 \sqrt{3}}+O\left(t^3\right)$$ $$\frac{1}{\sqrt{ 3 -2t}}=\frac{1}{\sqrt{3}}+\frac{t}{3 \sqrt{3}}+\frac{t^2}{6 \sqrt{3}}+O\left(t^3\right)$$ $$\frac{1+t}{\sqrt{ 3 -2t}}=\frac{1}{\sqrt{3}}+\frac{4 t}{3 \sqrt{3}}+O\left(t^2\right)$$ $$y=\frac{1}{\sqrt{3} \sqrt{t}}+\frac{4 \sqrt{t}}{3 \sqrt{3}}+O\left(t^{3/2}\right)$$ Now, replace $t$ by $\frac 1x$ to get $$y=\sqrt{\frac x 3}+\frac 4{2\sqrt{3x}}+\cdots$$ which shows the curve asymptote and how it is approached.

Edit

Concerning the different approximations, let us consider $$\Delta_1=\frac{x+1}{\sqrt{3x-2}}-\sqrt{\frac x 3}\qquad , \qquad \Delta_2=\frac{x+1}{\sqrt{3x-2}}-\frac{1}{3}\sqrt{3x-8}$$ and use Taylor expansion of the ratio. You should get $$\frac{\Delta_1}{\Delta_2}=\frac{1}{2}-\frac{7}{96 x}+O\left(\frac{1}{x^{3/2}}\right)$$ that is to say that you are twice better.

Answer 2

All of the asymptotes you have described are valid according to the definitions below:

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Similar to what is found here, we could say that $f(x)\sim g(x)$ means that $f$ and $g$ are relatively asymptotic: $$ f(x)\sim g(x)\implies \frac{f(x)}{g(x)}\to1\quad\text{as }x\to\infty $$ Based on this we could define a class of functions $[f(x)]^{\sim}$ consisting of all functions $g(x)$ being relatively asymptotic to $f$. This will be an equivalence class.

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Furthermore, we could define the notion of being strongly asymptotic, $f(x)\simeq g(x)$, as follows: $$ f(x)\simeq g(x)\implies |f(x)-g(x)|\to0\quad\text{as }x\to\infty $$ and the corresponding equivalence class $[f(x)]^{\simeq}$. The examples you gave are all in this latter class. For instance we have $$ \begin{align} \frac{x+1}{\sqrt{3x-2}}-\frac{\sqrt{x}}{\sqrt3} &=\frac{(x+1)\sqrt3-\sqrt{3x-2}\sqrt x}{\sqrt{3x-2}\sqrt 3}\\ &=\frac{(1+\frac1x)\sqrt3-\sqrt{3-\frac2x}}{\frac1x\sqrt{3x-2}\sqrt 3} \end{align} $$ and now since $\sqrt{3-t}>\sqrt3(1-\frac t3)$ for $t>0$ close enough to zero, which can be seen using that the derivative of $\sqrt x$ at $x=3$ is $\frac 1{2\sqrt 3}>\frac13$, we get with $t=\frac2x$ which is close to zero for large $x$ that $$ \begin{align} \frac{x+1}{\sqrt{3x-2}}-\frac{\sqrt{x}}{\sqrt3} &<\frac{(1+\frac1x)\sqrt3-\sqrt3(1-\frac2{3x})}{\frac1x\sqrt{3x-2}\sqrt 3}\\ &=\frac{\frac1{3x}}{\frac1x\sqrt{3x-2}}\\ &=\frac1{3\sqrt{3x-2}} \end{align} $$ which tends to zero as $x$ tends to infinity.