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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
    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 $\lim_{x\to 0}\frac{(2-\sqrt{4+x})(2+\sqrt{4+x})}{(2x\sqrt{4+x})(2+\sqrt{4+x})}$

  • 0
    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.