Another L'Hospital's Rule Issue

[ahorn]

I have another linear limit where multiplying by conjugates gives an indeterminate form of type \(\displaystyle \infty/\infty\) that can't get rid of by applying L'Hospital's Rule:

\(\displaystyle \displaystyle\lim_{x \to \infty}[\sqrt{9x^2+x}-3x]\)

If say \(\displaystyle \lim_{x \to \infty}\sqrt{9x^2+x-9x^2}=\infty\) my textbook tells me I'm wrong (but I don't know what rule I broke). The answer should be 1/6.
So I tried \(\displaystyle \lim_{x \to \infty}\frac{x}{\sqrt{9x^2 +x}+3x}=\lim_{x \to \infty}\frac{1}{\frac{18x+1}{2\sqrt{9x^2+x}}+3}\) but I can't get rid of the \(\displaystyle \frac{18x+1}{2\sqrt{9x^2+x}}\) when applying L'Hospital further to the denominator.

 

[lookagain]

Sorry, I should have tried multiplying by the conjugates. So the limit = 0?

I have another linear limit where multiplying by conjugates gives an indeterminate form of type \(\displaystyle \infty/\infty\) that can't get rid of by applying L'Hospital's Rule:

\(\displaystyle \displaystyle\lim_{x \to \infty}[\sqrt{9x^2+x}-3x]\)

If say \(\displaystyle \lim_{x \to \infty}\sqrt{9x^2+x-9x^2}=\infty \ \ \ \ \ \) <----- That doesn't equal the expression above.

my textbook tells me I'm wrong (but I don't know what rule I broke).


The answer should be 1/6.

So I tried \(\displaystyle \ \ \lim_{x \to \infty}\frac{x}{\sqrt{9x^2 +x}+3x}=\lim_{x \to \infty}\frac{1}{\frac{18x+1}{2\sqrt{9x^2+x}}+3} \ \ \ \ \ \) <----- No, do not try to use L'Hospital's Rule here.

but I can't get rid of the \(\displaystyle \frac{18x+1}{2\sqrt{9x^2+x}}\)

when applying L'Hospital further to the denominator.

You should start a separate thread for a different problem.


Here is a different strategy from that step on:


\(\displaystyle \ \ \ \displaystyle\lim_{x \to \infty}\frac{x}{\sqrt{9x^2 +x} \ + \ 3x} \ =\)



\(\displaystyle \ \ \displaystyle\lim_{x \to \infty}\frac{(1/x)x}{(1/x)\sqrt{9x^2 + x} \ + \ (1/x)3x} \ =\)



\(\displaystyle \ \ \displaystyle\lim_{x \to \infty}\frac{\frac{x}{x}}{\sqrt{\frac{9x^2}{x^2} + \frac{x}{x^2}} \ + \ \frac{3x}{x}} \ =\)



\(\displaystyle \ \ \displaystyle\lim_{x \to \infty}\frac{1}{\sqrt{9 + 1/x} \ + \ 3} \ =\)


Can you finish?

 

[ahorn]

I just figured it out, so I came to post that, but I see you beat me to it. \(\displaystyle \frac{1}{\sqrt{9+0}+3}=1/6\)

 

[HallsofIvy]

The "rule that you were breaking" is that \(\displaystyle \sqrt{9x^2+ x}- 3x\) is NOT equal to \(\displaystyle \sqrt{9x^2+ x- 9x^2}\).

In general, \(\displaystyle \sqrt{a- b}\ne \sqrt{a}- \sqrt{b}\).