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After I just learned there are ways to cancel out $0$s in the divisor of fractions onto limits I looked back at a task where I gave up when I got the result $\lim\limits_{x\to4} \frac {\sqrt {1+2x}-3} {\sqrt x -2}$ so I tried to find a way to get the $0$ out here, as well. Am I just not seeing the solution or is there no way to do it?

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  • $\begingroup$ A diiferent way than Arnaldo's is to use the general result that $\sqrt {1+y}=1+(y/2)(1+f(y))$ where $\lim_{y\to 0}f(x)=0.$......... So with $x=4+y$ we have $ -3+\sqrt {1+2x}=$ $\;-3+\sqrt {9+2y}=$ $\;-3+3\sqrt {1+2y/9}=$ $\;-3+3(1+(y/9)(1+f(2y/9))=$ $\;(y/3)(1+f(2y/9))$. And handle the denominator similarly. $\endgroup$ – DanielWainfleet Jan 24 '17 at 21:19
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Hint

$$\left(\frac{\sqrt{1+2x}-3}{\sqrt{x}-2}\cdot \frac{\sqrt{1+2x}+3}{\sqrt{x}+2}\right)\cdot \frac{\sqrt{x}+2}{\sqrt{1+2x}+3}=\frac{2x-8}{x-4}\cdot\frac{\sqrt{x}+2}{\sqrt{1+2x}+3}$$

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  • $\begingroup$ Thank you :) That's a nice trick $\endgroup$ – user405981 Jan 24 '17 at 19:53
  • $\begingroup$ you are very welcome! $\endgroup$ – Arnaldo Jan 24 '17 at 20:00
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With $\sqrt x=t+2$, the limit becomes

$$\lim_{t\to0}\frac{\sqrt{9+8t+2t^2}-3}t.$$

Then multiplying/dividing by the conjugate,

$$\lim_{t\to0}\frac{8t+t^2}t\frac1{\sqrt{9+8t+2t^2}+3}=8\frac1{3+3}.$$

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