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$$\lim_{x\to0} \frac{\sqrt{2x + 9} - 3}x$$

I tried to solve this but squaring both values but I cant get rid of the square root.

How to solve this?

Michael Hardy
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dfgj fghjk
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3 Answers3

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$$\frac{\sqrt{2x+9}-3}x=\frac{2x+9-9}{x(\sqrt{2x+9}+3)}=\frac2{\sqrt{2x+9}+3}\text{ if }x\ne0$$

Now we know if $x\to0,x\ne0$

Can you take it from here?

lab bhattacharjee
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    I'd call that more than just a "hint". – Michael Hardy Oct 04 '13 at 18:41
  • Ok, with that I got 2 / (sqrt (2x+9) +3). The sqrt is still there. – dfgj fghjk Oct 04 '13 at 18:43
  • @dfgj: Is that a problem? –  Oct 04 '13 at 18:44
  • So I should just replace the x with 0? Ok then I have 1/3. But i could have done that in the first place, right? Without using the formula you gave me, just replace x in the initial function with 0. I would have gotten dividing with 0, but still. – dfgj fghjk Oct 04 '13 at 18:47
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    @dfgj: This function is continuous at $0$ (do you know why?), so yes, you can replace $x$ with $0$. As you said, if you tried it on the original thing, you would have gotten dividing by zero, so you couldn't have done that in the first place. –  Oct 04 '13 at 18:50
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If you've gotten to differentiation yet, there's an "obvious" thing to do here.

The problem is that both $\sqrt{2x+9}$ and $3$ converge to the same value as $x \to 0$, and subtracting the two cancels out the most significant parts of their values.

So, we need to figure out the next most significant parts of their values: differential approximation (or more generally Taylor series) is an excellent way to do this.

Recall that differential approximation says that, e.g.,

$$ f(x) = f(0) + f'(0) x + x r(x) $$

where $r$ is a function that satisfies

$$ \lim_{x \to 0} r(x) = 0 $$

If you let $f(x) = \sqrt{2x+9}$, you can use differential approximation to rewrite your limit to make it simpler.

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The dominating element under the square root is the 9. If you factorize accordingly, $$ \frac{3\left(\sqrt{1+2x/9}-1\right)}{x} \sim \frac{3}{x}\times\frac{1}{2}\frac{2x}{9} = \frac{1}{3} $$

Siméon
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