Need some help finding this limit:


Here is what I have so far:




At this point I keep getting results I don't like, I have tried multiplying by the conjugate but I keep getting denominators of $0$. What am I missing here?



Another approach is to make the change of variables $y=\sqrt{x}$; because this is a continuous function, we know that the limit as $x\rightarrow 1$ is the same as the limit as $y\rightarrow 1$, so that we have $$\lim_{x\rightarrow 1}{ \frac{1-\sqrt{x}}{x\sqrt{x}-\sqrt{x}}}= \lim_{y\rightarrow 1}{\frac{1-y}{y^3-y}}=\lim_{y\rightarrow 1}{\frac{-(y-1)}{y(y-1)(y+1)}}=\lim_{y\rightarrow 1}{\frac{-1}{y(y+1)}}=-\frac{1}{2}$$


Hint: $1-x=\big(1-\sqrt x\big)\big(1+\sqrt x\big)$.


I find the route via L'Hôpital to be shortest here.

$$\lim_{x\rightarrow1}\frac{\frac{1}{\sqrt{x}}-1}{x-1}= \lim_{x\rightarrow1}\frac{\frac{d}{dx}(\frac{1}{\sqrt{x}}-1)}{\frac{d}{dx}(x-1)}= \lim_{x\rightarrow1}\;-\frac{1}{2}x^{-3/2}=-\frac{1}{2}$$

Same result. Different path.

A graph of the function shows the limit point at $-\frac{1}{2}$ as calculated. I would like to point out that the function is undefined at this point, but the graph fails to show this.

limit graph

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    $\begingroup$ That function is actually not defined at $x=1$, but is continuous on where it is defined. $\endgroup$ – user99914 Sep 12 '15 at 5:36
  • $\begingroup$ Thanks @John. I thought that if a function was not defined at a point, it would mean a discontinuity. Shouldn't the point (1, -1/2) be marked with an open circle? Please help me out here. :D $\endgroup$ – Adam Hrankowski Sep 12 '15 at 5:43
  • $\begingroup$ @JohnMa - Please see the discussion here: math.stackexchange.com/q/1431796/266049 $\endgroup$ – Adam Hrankowski Sep 12 '15 at 6:16
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    $\begingroup$ I guess you misunderstood my comment (from the discussion you linked). I did not say that $f$ is undefined and continuous at $x=1$. $\endgroup$ – user99914 Sep 12 '15 at 6:42
  • $\begingroup$ @JohnMa - Yes, I did misunderstand. I've now corrected the answer above. Thanks for your persistence. :) $\endgroup$ – Adam Hrankowski Sep 12 '15 at 14:25

$$\lim_{x\rightarrow 1}\frac{\frac{1}{\sqrt{x}}-1}{x-1}= \lim_{x\rightarrow 1}\frac{1-\sqrt{x}}{\sqrt{x}(x-1)}\cdot\frac{1+\sqrt{x}}{1+\sqrt{x}}=\lim_{x\rightarrow 1}\frac{1-x}{(x-1)\sqrt{x}(1+\sqrt{x})}=$$ $$=\lim_{x\rightarrow 1}-\frac{x-1}{\sqrt{x}(x-1)(1+\sqrt{x})}=\lim_{x\rightarrow 1}-\frac{1}{\sqrt{x}(1+\sqrt{x})}=-\frac{1}{2}. $$


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