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I saw this problem on a math board / insta:

$$\lim_{x\rightarrow3}\frac{\sqrt{3x}-3}{\sqrt{2x-4}-\sqrt{2}}$$

My first step would be take a derivative of the numerator and denominator to see if the limit exists or not, since just plugging in gets me 0/0 which is undefined.

$$\frac{d}{dx}(\sqrt{3x}-3)=\frac{d}{dx}((3x)^{\frac{1}{2}}-3)=\frac{3}{2}(3x)^{-1/2} $$

and

$$\frac{d}{dx}(\sqrt{2x-4}-\sqrt{2})=\frac{d}{dx}((2x-4)^{\frac{1}{2}}-\sqrt{2})=\frac{1}{2}(2x-4)^{-1/2}(2)=(2x-4)^{-1/2} $$

Which means at $x=3$ I get $\frac{3}{2}(9)^{1/2}= \frac{9}{2}$ in the numerator and $\sqrt{2}$ in the denominator to get me $$\lim_{x\rightarrow3}\frac{\sqrt{3x}-3}{\sqrt{2x-4}-\sqrt{2}}=\frac{9}{2\sqrt{2}}$$

So I am curious if I did this correctly, or if I am totally misusing L'Hôpital's rule.

ETA: thanks for noting the exponent error. So $\frac{3}{2}(3x)^{-1/2}$ at $x=3$ should be $\frac{3}{2\sqrt{3(3)}}=\frac{1}{2}$ and $(2x-4)^{-1/2}$ at $x=3$ is $\frac{1}{\sqrt{2}}$

for $\frac{\sqrt{2}}{2}$

Jesse
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2 Answers2

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As noticed there is a mistake with a negative exponent, as an alternative by rationalization

$$\frac{\sqrt{3x}-3}{\sqrt{2x-4}-\sqrt{2}}=\frac{\sqrt{3x}-3}{\sqrt{2x-4}-\sqrt{2}}\frac{\sqrt{3x}+3}{\sqrt{3x}+3}\frac{\sqrt{2x-4}+\sqrt{2}}{\sqrt{2x-4}+\sqrt{2}}=$$

$$=\frac{3x-9}{2x-6}\frac{\sqrt{2x-4}+\sqrt{2}}{\sqrt{3x}+3}=\frac{3}{2}\frac{\sqrt{2x-4}+\sqrt{2}}{\sqrt{3x}+3} \to\frac32\frac{2\sqrt 2}{6}=\frac{\sqrt{2}}2$$

Refer also to the related

user
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  • In the second row, I believe the second fraction is upside down. And then that propagates. – 2'5 9'2 Oct 10 '20 at 21:38
  • @alex.jordan Yes of course! Thanks I fix that – user Oct 10 '20 at 21:42
  • And the actual limit is $\sqrt{2}/2$. But the mistake here coincidentally agrees with the mistake in OPs math where they did not use negative exponents correctly. – 2'5 9'2 Oct 10 '20 at 21:42
  • @alex.jordan Indeed this was way I didn't noticed that at first! Thanks – user Oct 10 '20 at 21:44
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Uhm, you get $2^{-1/2}$ in the denominator, not $2^{1/2}=\sqrt{2}$.

Let's try in a different way: first $$ \lim_{x\to3}\frac{\sqrt{3x}-3}{x-3}= \lim_{x\to3}\frac{3x-9}{x-3}\frac{1}{\sqrt{3x}+3}=\frac{3}{6}=\frac{1}{2} $$ Also $$ \lim_{x\to3}\frac{\sqrt{2x-4}-\sqrt{2}}{x-3}=\lim_{x\to3}\frac{2x-4-2}{x-3}\frac{1}{\sqrt{2x-4}+\sqrt{2}}=\frac{2}{2\sqrt{2}}=\frac{1}{\sqrt{2}} $$ Hence your limit is $$ \lim_{x\to3}\frac{\sqrt{3x}-3}{\sqrt{2x-4}-\sqrt{2}} =\lim_{x\to3}\frac{\sqrt{3x}-3}{x-3}\,\frac{x-3}{\sqrt{2x-4}-\sqrt{2}} =\frac{1}{2}\sqrt{2}=\frac{1}{\sqrt{2}} $$ What's going wrong?

The derivative of $f(x)=\sqrt{3x}-3$ is $$ f'(x)=\frac{3}{2\sqrt{3x}} $$ and therefore $f'(3)=3/6=1/2$.

The derivative of $g(x)=\sqrt{2x-4}-\sqrt{2}$ is $$ g'(x)=\frac{2}{2\sqrt{2x-4}} $$ and $g'(3)=1/\sqrt{2}$.

You seem to confuse $a^{-1/2}$ with $a^{1/2}$.

egreg
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  • I saw the mistakes I made with the negative exponents, but I am not sure the motivation for what you have here- the first step is a bit opaque to me as I wasn't sure why you chose x-3 as the denominator there, though the rationalization via $\sqrt{3x}+3$ makes a bit more sense... – Jesse Oct 11 '20 at 17:06
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    @Jesse The idea is to try and write $\frac{f(x)}{g(x)}$ as the quotient of $\frac{f(x)}{x-3}$ by $\frac{g(x)}{x-3}$ to see whether this simplifies the derivation, doing just one rationalization in each case. It does, in this particular case, essentially because a single application of l'Hôpital is needed. It reduces the risk of making mistakes. – egreg Oct 11 '20 at 17:10
  • OK, so why x-3 tho? The radical 3x seems to make that less obvious to me. – Jesse Oct 11 '20 at 17:23
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    @Jesse We want an indeterminate form, don't we? Since both $f(x)=\sqrt{3x}-3$ and $g(x)=\sqrt{2x-4}-\sqrt{2}$ have limit zero for $x\to 3$, we get indeterminate forms (actually the ratios in the definition of derivative) when dividing by $x-3$. – egreg Oct 11 '20 at 17:28