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This question already has an answer here:

Find the following limit: \begin{equation*} \lim_{x \rightarrow 4} \frac{\sqrt{1 + 2x} -3}{\sqrt{x} - 2} \end{equation*}

I have tried to divide the numerator and denominator by $\sqrt{x}$, but it did not work.

I have tried to multiply by the conjugates of the numerator and denominator simultaneously but it did not work.

I have tried to multiply by the conjugates of the numerator only but it did not work.

So what shall I do?

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marked as duplicate by Nosrati, Paramanand Singh calculus Oct 2 '18 at 7:27

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Why did your 2nd trial fail? That is viable actually. $\endgroup$ – xbh Oct 2 '18 at 4:01
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Hint:$$\frac{\sqrt{1+2x}-3}{\sqrt{x}-2} = \frac{1+2x-9}{(\sqrt{x}-2)(\sqrt{1+2x}+3)} = \frac{2(\sqrt{x}-2)(\sqrt{x}+2)}{(\sqrt{x}-2)(\sqrt{1+2x}+3)}$$

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Using a substitution and a derivative:

Set $x=t^2$: $$ \lim_{x \rightarrow 4} \frac{\sqrt{1 + 2x} -3}{\sqrt{x} - 2} = \lim_{t \rightarrow 2} \frac{\sqrt{1 + 2t^2} -3}{t - 2} = f'(2) \mbox{ for } f(t) = \sqrt{1 + 2t^2}$$

So, $$f'(t) = \frac{2t}{\sqrt{1 + 2t^2}} \Rightarrow \lim_{x \rightarrow 4} \frac{\sqrt{1 + 2x} -3}{\sqrt{x} - 2} = f'(2) = \frac{4}{3}$$

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Rationalise the numerator.

$\lim_{x\rightarrow4}\left(\dfrac{\sqrt{1+2x}-3}{\sqrt{x}-2}\right)=\lim_{x\rightarrow4}\left(\dfrac{\sqrt{1+2x}-3}{\sqrt{x}-2}\times\dfrac{\sqrt{1+2x}+3}{\sqrt{1+2x}+3}\right)=\lim_{x\rightarrow4}\left(\dfrac{2(\sqrt{x}+2)^2}{(\sqrt{x}-2)(\sqrt{1+2x}+3)}\right)=\dfrac{2(\sqrt{4}+2)}{\sqrt{1+2(4)}+3}=\dfrac{8}{9}=\dfrac43$

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  • $\begingroup$ $(\sqrt{1 + 2x} - 3)(\sqrt{1 + 2x}+ 3) = 1 + 2x - 9 = 2x - 8 = 2(x - 4) = 2(\sqrt{x} + 2)(\sqrt{x} - 2)$, which is why you can cancel a factor of $\sqrt{x} - 2$. $\endgroup$ – N. F. Taussig Oct 2 '18 at 10:23
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If you want to go even beyond the limit itself, let $x=y+4$ to make $$A=\frac{\sqrt{2 x+1}-3}{\sqrt{x}-2}=\frac{\sqrt{2 y+9}-3}{\sqrt{y+4}-2}$$ and use the binomial expansion or (better) Taylor series around $y=0$. This would give $$A=\frac{\frac{y}{3}-\frac{y^2}{54}+O\left(y^3\right) }{\frac{y}{4}-\frac{y^2}{64}+O\left(y^3\right)}=\frac{4}{3}+\frac{y}{108}+O\left(y^2\right)$$ Back to $x$, $$A=\frac{4}{3}+\frac{x-4}{108}+O\left((x-4)^2\right)$$

Try using $x=5$; the exact value is $A=\left(2+\sqrt{5}\right) \left(\sqrt{11}-3\right)\approx 1.34124$ while the expansion gives $\frac{145}{108}\approx 1.34259$.

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