l'Hopital's Rule: lim [->infty] [root x^2 + 3x - x]

[thelazyman]

I am doing a question using l'hopitals rule and I don't know how to do full solutions to this problem. (The math teacher only gave us an answer.) I got most of it, but I didn't get the last step.

The question is:

. . .lim-->infinity ( root x^2 + 3x - x) = lim --> infinity x(root 1 + 3/x - 1

Then I get root 1 + 3x -1 over 1/x.

Then get from deriving 1/2 (1+ 3/x)^-1/2 (-3/x^2) over -1/x^2

Dont know what they do to factor the rest to get 3/2(1/root 1 + 3x) = 3/2

Dont know the factors for the rest. Please help

 

[tkhunny]

??? This makes very little sense. l'hopital's rule requires something called an "Indeterminate Form". You do NOT have such a thing. You must first demonstrate that the rule is applicable, THEN you can use it. Demonstrate away.

Note: Why would anyone be motivated to answer questions for someone known as "thelazyman"?

Note2: That can't be typed correctly. Why would there be two 'x'-terms in the original problem statement.

 

[soroban]

Re: l'hopitals rule

Hello, thelazyman!

Some parentheses would have helped . . .

\(\displaystyle \lim_{x\to\infty}\,\left[\sqrt{x^2\,+\,3x}\,- \,x\right]\)

This goes to the indeterminate form: \(\displaystyle \,\infty\,-\,\infty\)

To use L'Hopital, the form must be: \(\displaystyle \:\frac{0}{0}\) or \(\displaystyle \frac{\infty}{\infty}\)


Multiply top and bottom by \(\displaystyle \left(\sqrt{x^2\,+\,3x}\,+\,x\right)\)

. . \(\displaystyle \L\frac{\sqrt{x^2\,+\,3x}\,-\,x}{1}\,\cdot\,\frac{\sqrt{x^2\,+\,3x}\,+\,x}{\sqrt{x^2\,+\,3x}\,+\,x} \;=\;\frac{(x^2\,+\,3x)\,-\,x^2}{\sqrt{x^2\,+\,3x}\,+\,x} \;=\;\frac{3x}{\sqrt{x^2\,+\,3x}\,+\,x}\)

. . and this form goes to \(\displaystyle \frac{\infty}{\infty}\) . . . We can use L'Hopital


We have: \(\displaystyle \L\:\frac{3x}{(x^2\,+\,3x)^{\frac{1}{2}}\,+\,x}\)

Using L'Hopital: \(\displaystyle \L\:\frac{3}{\frac{1}{2}\left(x^2\,+\,3x\right)^{-\frac{1}{2}}(2x\,+\,3)\,+\,1}\;=\;\frac{3}{\frac{2x\,+\,3}{2\sqrt{x^2\,+\,3x}}\,+\,1}\)

Good luck!