End Behaviour of ƒ(x) = √(x2+x) - x: Calculating Asymptotes with Derivatives

[Mathematically]

Homework Statement

Determine the end behaviour of the function ƒ(x) = √(x²+x) - x

Homework Equations

Not quite sure, but I suppose f'(x) would be useful?

The Attempt at a Solution

Essentially what I did was take the derivative of f(x), however I was unsure as to how one would take the limit of the derivative as x approaches positive or negative infinity (to see if f(x) has any asymptotes). Thanks in advance for any help, sorry if I explained this poorly.

 

[BvU]

Hello MM,

Thanks for introducing this "end behaviour" term to me; I didn't know about it .
I suppose you are interested in limits to $\pm\infty$ .

Homework Equations

Not quite sure, but I suppose f'(x) would be useful?

Not exactly an equation, is it ? What does it give you when $x\rightarrow +\infty$ ?

I suppose you are interested in limits to $\pm\infty$ .
For $+\infty$, how can you rewrite f(x) ?

 

[Ray Vickson]

Homework Statement

Determine the end behaviour of the function ƒ(x) = √(x²+x) - x

Homework Equations

Not quite sure, but I suppose f'(x) would be useful?

The Attempt at a Solution

Essentially what I did was take the derivative of f(x), however I was unsure as to how one would take the limit of the derivative as x approaches positive or negative infinity (to see if f(x) has any asymptotes). Thanks in advance for any help, sorry if I explained this poorly.

$$\sqrt{x^2+x}-x = \sqrt{x^2} \left(1 + \frac{1}{x} \right)^{1/2} -x = |x| \left(1 + \frac{1}{x} \right)^{1/2} -x $$
What happens to $(1+z)^{1/2}$ for small $z = 1/x$?

 

[Mark44]

Homework Statement

Determine the end behaviour of the function ƒ(x) = √(x²+x) - x

Homework Equations

Not quite sure, but I suppose f'(x) would be useful?

Not really. For the end behavior, all you need to do is to evaluate the limits $\lim_{x \to \infty}f(x)$ and $\lim_{x \to -\infty}f(x)$.

The Attempt at a Solution

Essentially what I did was take the derivative of f(x), however I was unsure as to how one would take the limit of the derivative as x approaches positive or negative infinity (to see if f(x) has any asymptotes). Thanks in advance for any help, sorry if I explained this poorly.

 

[Mathematically]

$$\sqrt{x^2+x}-x = \sqrt{x^2} \left(1 + \frac{1}{x} \right)^{1/2} -x = |x| \left(1 + \frac{1}{x} \right)^{1/2} -x $$
What happens to $(1+z)^{1/2}$ for small $z = 1/x$?

So as x approaches infinity, will (1+1/x)^(1/2) approach 1? Thus, the function approaches |x| - x ?

 

[Ray Vickson]

So as x approaches infinity, will (1+1/x)^(1/2) approach 1? Thus, the function approaches |x| - x ?

Yes, as $x \to +\infty$ we have $(1+ 1/x)^{1/2} \to 1$, but for large $x>0$ (not yet at $x = \infty$) there will be some small correction terms; that is,
$$\left(1 + \frac{1}{x}\right)^{1/2} = 1 + \text{small correction terms}$$
for large $x$. Don't forget that we multiply $(1 + 1/x)^{1/2}$ by $|x|$, so we multiply the small correction terms by the large value of $|x|$. The result of doing that will be something that does NOT go to zero as $x$ goes to $\infty$. You need to figure out what the main part of the small correction term actually is, in order to arrive at a correct limiting behavior. If you did not do that you would arrive at $\lim_{x \to \infty} \: x \cdot 1 - x = 0$, and that is not the right answer.

 

[Mathematically]

Yes, as $x \to +\infty$ we have $(1+ 1/x)^{1/2} \to 1$, but for large $x>0$ (not yet at $x = \infty$) there will be some small correction terms; that is,
$$\left(1 + \frac{1}{x}\right)^{1/2} = 1 + \text{small correction terms}$$
for large $x$. Don't forget that we multiply $(1 + 1/x)^{1/2}$ by $|x|$, so we multiply the small correction terms by the large value of $|x|$. The result of doing that will be something that does NOT go to zero as $x$ goes to $\infty$. You need to figure out what the main part of the small correction term actually is, in order to arrive at a correct limiting behavior. If you did not do that you would arrive at $\lim_{x \to \infty} \: x \cdot 1 - x = 0$, and that is not the right answer.

Ah, I see. That makes sense although I am not sure how to actually go about doing that; my knowledge of calculus is very very limited. I think my teacher is showing us his solution soon so hopefully it will become clear. Thanks so much for the help!

 

[BvU]

If you want to do some reconnoitering: Do sqrt(1.01) , sqrt(1.001) sqrt(1.0001) etc on your calculator. Or peek at the Taylor series for $\ \sqrt{1+\epsilon}\ \ \$.