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$\begin{align*} \lim_{x\to 0}\frac{\frac{1}{\sqrt{4+x}}-\frac{1}{2}}{x} &=\lim_{x\to 0}\frac{\frac{2}{2\sqrt{4+x}}-\frac{\sqrt{4+x}}{2\sqrt{4+x}}}{x}\\ &=\lim_{x\to 0}\frac{\frac{2-\sqrt{4+x}}{2\sqrt{4+x}}}{x}\\ &=\lim_{x\to 0}\frac{2-\sqrt{4+x}}{2x\sqrt{4+x}}\\ &=\lim_{x\to 0}\frac{(2-\sqrt{4-x})(2+\sqrt{4-x})}{(2x\sqrt{4+x})(2+\sqrt{4-x})}\\ &=\lim_{x\to 0}\frac{2 \times 2 + 2\sqrt{4-x}-2\sqrt{4-x}-((\sqrt{4-x})(\sqrt{4-x})) }{2 \times 2x\sqrt{4+x} + 2x\sqrt{4+x}\sqrt{4-x}}\\ &=\lim_{x\to 0}\frac{4-4+x}{4x\sqrt{4+x} + 2x\sqrt{4+x}\sqrt{4-x}}\\ &=\lim_{x\to 0}\frac{x}{x(4\sqrt{4+x} + 2\sqrt{4+x}\sqrt{4-x})}\\ &=\lim_{x\to 0}\frac{1}{(4\sqrt{4+x} + 2\sqrt{4+x}\sqrt{4-x})}\\ &=\frac{1}{(4\sqrt{4+0} + 2\sqrt{4+0}\sqrt{4-0})}\\ &=\frac{1}{16} \end{align*}$

wolframalpha says it's negative. What am I doing wrong?

  • 0
    @Asaf: The correct place for the equal sign (or for an operation sign if you were to continue on the next line) is at the beginning of the new line, not the end of the old one.2011-02-18
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    You swapped radicals from $\sqrt{4+x}$ to $\sqrt{4-x}$ somewhere along the line. The change of sign messes everything up afterwards.2011-02-18
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    @Arturo: Duly noted! Thanks!2011-02-18
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    It is the derivative of $\frac{1}{\sqrt{4+x}}$ at $0$. Since this is a decreasing function for $x \gt 0$ and the derivative is continuous, it follows that the limit you seek is non-positive at $0$.2011-02-18
  • 0
    Performing less calculations reduces chance for errors: third binomial formula had not saved your day, but might another.2011-02-18

4 Answers 4

15

Others have already pointed out a sign error. One way to avoid such is to first simplify the problem by changing variables. Let $\rm\ z = \sqrt{4+x}\ $ so $\rm\ x = z^2 - 4\:.\:$ Then

$$\rm \frac{\frac{1}{\sqrt{4+x}}-\frac{1}{2}}{x}\ =\ \frac{\frac{1}z - \frac{1}2}{z^2-4}\ =\ \frac{-(z-2)}{2\:z\:(z^2-4)}\ =\ \frac{-1}{2\:z\:(z+2)}$$

In this form it is very easy to compute the limit as $\rm\ z\to 2\:$.

1

Certainly for $x \gt 0,\frac{1}{\sqrt{4+x}}-\frac{1}{2} \lt 0$ so the limit should be negative. Between the fifth and sixth limit you flipped a sign under the sqrt in the numerator and that changes the sign of the total thing

1

In between the fourth and fifth steps, you go from $$\lim_{x\to 0}\frac{2-\sqrt{4+x}}{2x\sqrt{4+x}} \text{ to } \lim_{x\to 0}\frac{(2-\sqrt{4-x})(2+\sqrt{4-x})}{(2x\sqrt{4+x})(2+\sqrt{4-x})}$$ which is not correct. It should be $$\lim_{x\to 0}\frac{(2-\sqrt{4+x})(2+\sqrt{4+x})}{(2x\sqrt{4+x})(2+\sqrt{4+x})}$$

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    You've retained the incorrect sign on $x$ for the conjugate. All the radical terms should be $\sqrt{4+x}$. (What you wrote isn't actually incorrect, it just wouldn't help as much.)2011-02-18
0

An argument why it should be negative.

When $x>0$, we have $\sqrt{4+x} > 2$ and hence $\frac{1}{\sqrt{4+x}} < \frac{1}{2}$ and hence

$$\frac{\frac{1}{\sqrt{4+x}} - \frac{1}{2}}{x} < 0$$

Similarly, When $x<0$, we have $\sqrt{4+x} < 2$ and hence $\frac{1}{\sqrt{4+x}} > \frac{1}{2}$ and hence

$$\frac{\frac{1}{\sqrt{4+x}} - \frac{1}{2}}{x} < 0$$

So either way, as $x \rightarrow 0$, the value is negative.