# Find the limit of $\begin{equation*} \lim_{x \rightarrow 4} \frac{\sqrt{1 + 2x} -3}{\sqrt{x} - 2} \end{equation*}$ [duplicate]

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?

## marked as duplicate by Nosrati, Paramanand Singh calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 2 '18 at 7:27

• Why did your 2nd trial fail? That is viable actually. – xbh Oct 2 '18 at 4:01

Hint:$$\frac{\sqrt{1+2x}-3}{\sqrt{x}-2} = \frac{1+2x-9}{(\sqrt{x}-2)(\sqrt{1+2x}+3)} = \frac{2(\sqrt{x}-2)(\sqrt{x}+2)}{(\sqrt{x}-2)(\sqrt{1+2x}+3)}$$

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

Rationalise the numerator.

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

• $(\sqrt{1 + 2x} - 3)(\sqrt{1 + 2x}+ 3) = 1 + 2x - 9 = 2x - 8 = 2(x - 4) = 2(\sqrt{x} + 2)(\sqrt{x} - 2)$, which is why you can cancel a factor of $\sqrt{x} - 2$. – N. F. Taussig Oct 2 '18 at 10:23

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 $$\frac{145}{108}\approx 1.34259$$.