evaluating limits algebraically: limit [x->0] [(1/sqrt{9 + x} - 1/3) / x]

[thecousinleo]

\[\lim_{x \rightarrow 0}\frac{\frac{1}{\sqrt{9+x}}-\frac{1}{3}}{x}\]
Can anyone solve this, please? I'm stuck!

 

[tkhunny]

What's stopping you from adding those two terms in the numerator?

 

[Dr.Peterson]

Two tools that will help are common denominators, and conjugates. Once you show us where you are stuck, we will have a better idea how to help.

 

[mmm4444bot]

Please read the forum's submission guidelines. You can start with this summary. Thank you!

?

 

[thecousinleo]

I'm sorry! I did that (common denominators and conjugates). And I got here:
\[\lim_{x \rightarrow 0}\frac{\frac{3-\sqrt{9+x}}{3\sqrt{9+x}}}{x}
\lim_{x \rightarrow 0}\frac{3-\sqrt{9+x}}{3x\sqrt{9+x}}.\frac{\sqrt{9+x}}{\sqrt{9+x}}\lim_{x \rightarrow 0}\frac{\sqrt{9+x}(3-\sqrt{9+x})}{3x(\sqrt{9+x})^{2}}\lim_{x \rightarrow 0}\frac{\sqrt{9+x}(3-\sqrt{9+x})}{3x(9+x)}\]
But I still get 0/0 if I plug the zero, and I can't see what else I can do!

 

[Dr.Peterson]

I don't see any conjugates there! You need to multiply by [MATH]\frac{3+\sqrt{9+x}}{3+\sqrt{9+x}}[/MATH], not [MATH]\frac{\sqrt{9+x}}{\sqrt{9+x}}[/MATH].

 

[thecousinleo]

Oh, I see, my mystake, sorry! Need to look up the real meaning for conjugates, thnks!
So:
\[\lim_{x \rightarrow 0}\frac{(3-\sqrt{9+x})}{(3x\sqrt{9+x})}.\frac{(3+\sqrt{9+x})}{(3+\sqrt{9+x})}\lim_{x \rightarrow 0}\frac{9-(9+x)}{(3x\sqrt{9+x})(3+\sqrt{9+x})}\lim_{x \rightarrow 0}\frac{-x}{(3x\sqrt{9+x})(3+\sqrt{9+x})}\]
Then I'll cancel the X, and 'll have as a result:
\[\lim_{x \rightarrow 0}-\frac{1}{(3\sqrt{9+x})(3+\sqrt{9+x)}}=-\frac{1}{54}\]
Is that correct?

 

[Steven G]

It looks good. You can check it yourself by plugging in a value for x in this case near 0, say .0001. Get an answer and subtract -1/54 from your answer and if you get close to 0, then you are (probably) correct.

As far as conjugate you need to know that (a+b)(a-b) = a^2 - b^2

 

[lookagain]

\[\lim_{x \rightarrow 0}\frac{\frac{1}{\sqrt{9+x}}-\frac{1}{3}}{x}\]
Can anyone solve this, please? I'm stuck!

You want to avoid writing/typing radical symbols (read: time consuming and prone to error), using the conjugates,
and combining fractions as an option.

Let \(\displaystyle \ \sqrt{9 + x} = u\)

Then \(\displaystyle \ 9 + x = u^2\).

And \(\displaystyle \ x = u^2 - 9\).

As x approaches 0, u approaches 3.

Substitute:

\(\displaystyle \displaystyle\lim_{u \to 3}\dfrac{1/u \ - \ 1/3}{u^2 - 9}\)

\(\displaystyle \displaystyle\lim_{u \to 3}\dfrac{3u(1/u \ - \ 1/3)}{3u(u^2 - 9)}\)

\(\displaystyle \displaystyle\lim_{u \to 3}\dfrac{ 3- u}{3u(u - 3)(u + 3)}\)

\(\displaystyle \displaystyle\lim_{u \to 3}\dfrac{-1(u - 3)}{3u(u - 3)(u + 3)}\)

Continue from there.

 

[thecousinleo]

You want to avoid writing/typing radical symbols (read: time consuming and prone to error), using the conjugates,
and combining fractions as an option.

Let \(\displaystyle \ \sqrt{9 + x} = u\)

Then \(\displaystyle \ 9 + x = u^2\).

And \(\displaystyle \ x = u^2 - 9\).

As x approaches 0, u approaches 3.

Substitute:

\(\displaystyle \displaystyle\lim_{u \to 3}\dfrac{1/u \ - \ 1/3}{u^2 - 9}\)

\(\displaystyle \displaystyle\lim_{u \to 3}\dfrac{3u(1/u \ - \ 1/3)}{3u(u^2 - 9)}\)

\(\displaystyle \displaystyle\lim_{u \to 3}\dfrac{ 3- u}{3u(u - 3)(u + 3)}\)

\(\displaystyle \displaystyle\lim_{u \to 3}\dfrac{-1(u - 3)}{3u(u - 3)(u + 3)}\)

Continue from there.

I'm sorry, I can't see why u approaches 3. Could you explain that, please?

 

[tkhunny]

Definition? [math]\sqrt{9+x} = u\;and\;x \rightarrow 0[/math]