I need this limit ((x(sqrt(x+2)))/sqrt(x+1))-x to calculate the asymptote of this function: ((x(sqrt(x+2)))/sqrt(x+1)).

which, according to the class notes: y=x+1/2

with a = 1, b = 1/2

However, the online math calculators say that currently no steps are supported to show for this kind of problem.

I calculated this limit as x*sqrt(1)-x = 1, but apparently the correct answer is 1/2.

What is my mistake?

Thx in advance.

r

[latex]\frac{x(\sqrt{x+2})}{\sqrt{x+1}}-x=\frac{x}{\sqrt{x+1}}(\sqrt{x+2}-\sqrt{x+1})=

\frac{x}{\sqrt{x+1}(\sqrt{x+2}+\sqrt{x+1})}(x+2-(x+1))[/latex]

Limit = 1/2, since the denominator ->2x.

You should use l'Hôpital's rule for a rigorous proof:
[math]\lim_{x\to\infty} {\frac{x\sqrt{x+2}}{\sqrt{x+1}}-x}=\lim_{x\to\infty} {\frac{x}{\sqrt{(x+1)(x+2)}+x+1}}[/math] (see the previous post).

Using l'Hôpital's rule, this equals [math]\lim_{x\to\infty} {\frac{1}{(\sqrt{(x+1)(x+2)})'+1}}=\lim_{x\to\infty} {\frac{1}{\frac{2x+3}{2\sqrt{(x+1)(x+2)}}+1}}[/math]. Assuming the limit exists, this is equal to [math]\frac{1}{\lim_{x\to\infty} {\frac{2x+3}{2\sqrt{(x+1)(x+2)}}}+1}[/math].

You then evaluate [math]\lim_{x\to\infty} {\frac{2x+3}{2\sqrt{(x+1)(x+2)}}}[/math]: Using l'Hôpital again, this is equal to [math]\lim_{x\to\infty} {\frac{2}{\frac{2(2x+3)}{2\sqrt{(x+1)(x+2)}}}}=\lim_{x\to\infty} {\frac{2\sqrt{(x+1)(x+2)}}{2x+3}}[/math]. It follows that this limit is 1, and therefore [math]\lim_{x\to\infty} {\frac{x\sqrt{x+2}}{\sqrt{x+1}}-x}=\frac{1}{1+1}=\frac{1}{2}[/math].

Edited September 28, 20169 yr by renerpho

Why use L'Hopital's rule? Mathematic's answer is much simpler.

You can use wolframalpha to visualize the function, fyi:

 

https://www.wolframalpha.com/input/?i=((x(sqrt(x%2B2)))%2Fsqrt(x%2B1))-x