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Need some help finding this limit:

$$\lim_{x\rightarrow1}\frac{\frac{1}{\sqrt{x}}-1}{x-1}$$

Here is what I have so far:

$$\lim_{x\rightarrow1}\dfrac{\dfrac{1-\sqrt{x}}{\sqrt{x}}}{x-1}$$

$$\lim_{x\rightarrow1}\dfrac{1-\sqrt{x}}{\sqrt{x}}\cdot\dfrac{1}{x-1}$$

$$\lim_{x\rightarrow1}\dfrac{1-\sqrt{x}}{x\sqrt{x}-\sqrt{x}}$$

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?

Thanks

alvonellos
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nitrous2
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4 Answers4

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

Lucian
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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 $$\lim_{x\rightarrow 1}{ \frac{1-\sqrt{x}}{x\sqrt{x}-\sqrt{x}}}= \lim_{y\rightarrow 1}{\frac{1-y}{y^3-y}}=\lim_{y\rightarrow 1}{\frac{-(y-1)}{y(y-1)(y+1)}}=\lim_{y\rightarrow 1}{\frac{-1}{y(y+1)}}=-\frac{1}{2}$$

Hayden
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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

MathAdam
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    That function is actually not defined at $x=1$, but is continuous on where it is defined. –  Sep 12 '15 at 05:36
  • 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 – MathAdam Sep 12 '15 at 05:43
  • @JohnMa - Please see the discussion here: http://math.stackexchange.com/q/1431796/266049 – MathAdam Sep 12 '15 at 06:16
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    I guess you misunderstood my comment (from the discussion you linked). I did not say that $f$ is undefined and continuous at $x=1$. –  Sep 12 '15 at 06:42
  • @JohnMa - Yes, I did misunderstand. I've now corrected the answer above. Thanks for your persistence. :) – MathAdam Sep 12 '15 at 14:25
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$$\lim_{x\rightarrow 1}\frac{\frac{1}{\sqrt{x}}-1}{x-1}= \lim_{x\rightarrow 1}\frac{1-\sqrt{x}}{\sqrt{x}(x-1)}\cdot\frac{1+\sqrt{x}}{1+\sqrt{x}}=\lim_{x\rightarrow 1}\frac{1-x}{(x-1)\sqrt{x}(1+\sqrt{x})}=$$ $$=\lim_{x\rightarrow 1}-\frac{x-1}{\sqrt{x}(x-1)(1+\sqrt{x})}=\lim_{x\rightarrow 1}-\frac{1}{\sqrt{x}(1+\sqrt{x})}=-\frac{1}{2}. $$

ZHN
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