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How do I evaluate the limit of $$\lim_{x\to 0}\frac{\sqrt{x+1}-1}{x} = \frac{1}{2}$$? As $x$ approaches $0$, I know the answer is $\frac{1}{2}$, but I got this question wrong. I think you have to multiply by the conjugate of the numerator? Can someone explain the steps to this, I must be doing something wrong when I multiply.

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    $\begingroup$ Hint: Recall the definition of a derivative. $\endgroup$ – Andrew D Sep 10 '13 at 17:47
  • $\begingroup$ For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$ – Martin Sleziak Sep 10 '13 at 17:50
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    $\begingroup$ The problem with referring to the limit definition of derivative is that this is a sort of problem that appears in an introductory course well before derivatives are defined. (Afterwards, naturally, it is clearer what can be done with this limit...) $\endgroup$ – colormegone Sep 10 '13 at 17:52
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$\require{cancel}$ Multiply numerator and denominator by the conjugate of the numerator: $$\sqrt{x+1} + 1$$ then evaluate the limit.

When we multiply by the conjugate, recall how we factor the difference of squares: $$(\sqrt a - b) \cdot (\sqrt a + b) = (\sqrt a)^2 - b^2 = a - b^2$$

$$\dfrac {\sqrt{x+1} - 1}{x} \cdot \frac{\sqrt{x+1} + 1}{\sqrt{x+1} + 1} = \dfrac {(x + 1) - 1 }{x(\sqrt{x+1} + 1)}= \dfrac {\cancel{x}}{\cancel{x}(\sqrt{x+1} + 1)} = \dfrac 1{\sqrt{x + 1} + 1}$$

Now we need only to evaluate $$\lim_{x \to 0} \dfrac 1{\sqrt{x + 1} + 1}$$

I trust you can do that.

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  • $\begingroup$ what do you get once you multiply? I think I'm doing my math wrong when I multiply the numerator by the conjugate. $\endgroup$ – kb95825 Sep 10 '13 at 17:50
  • $\begingroup$ Recall the difference of squares: the numerator becomes: $$(x+1) - 1 = x$$ $\endgroup$ – Namaste Sep 10 '13 at 17:51
  • $\begingroup$ @amWhy I suggest you fix the braces in the remainders ;-) $\endgroup$ – AlexR Sep 10 '13 at 17:57
  • $\begingroup$ @kb95825 Recall that we need to multiply numerator and denominator by the conjugate of the numerator: that is like multiplying by $1$, so it doesn't change the original function. $\endgroup$ – Namaste Sep 10 '13 at 17:57
  • $\begingroup$ yes thank you :) $\endgroup$ – kb95825 Sep 10 '13 at 21:25
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Method 1 (basic)
$$\frac{\sqrt{x+1} - 1}{x} \stackrel{\sqrt{x+1}^2 = |x+1|}{=} \frac{|x + 1| - 1}{x (\sqrt{x+1} + 1)} \stackrel{x+1 \geq 0, \text{ for well-def.}}{=} \frac{1}{\sqrt{x + 1} + 1} \to \frac{1}{1+1} = \frac{1}{2}$$ as $x\to 0$, because $x\mapsto \sqrt{x}$ is continuous (and so the limit can be "used as input")
Method 2 (derivative)
$$\frac{\sqrt{x+1} - 1}{x} = \frac{f(1 + x) - f(1)}{x}$$ where $f(y) = \sqrt{y}$, thus the limit is $$f'(1) = \frac{1}{2 \sqrt{1}} = \frac{1}{2}$$ Method 3 (l'Hospital)
Since the form is $\frac{0}{0}$, l'Hospital can be applied and gives $$\lim_{x\to 0} \frac{\sqrt{x+1} - 1}{x} = \lim_{x\to 0} \frac{\frac{1}{2\sqrt{x+1}}}{1} = \lim_{x\to 0} \frac{1}{2\sqrt{x + 1}} = \frac{1}{2 \sqrt{1}} = \frac{1}{2}$$

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  • $\begingroup$ Perhaps it is also worth mentioning in method 1 that $(\sqrt{x+1})^2=|x+1|$, but since the expression in question is only defined for $x+1\ge0$, we can replace $|x+1|$ by $(x+1)$. $\endgroup$ – Martin Sleziak Sep 10 '13 at 17:54
  • $\begingroup$ @MartinSleziak Yep, that's true. I'll add it $\endgroup$ – AlexR Sep 10 '13 at 17:54
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Alternative solution: Write $ t= \sqrt{x+1}$ then $x=t^2-1$ and we get $$\lim_{x\to 0}\frac{\sqrt{x+1}-1}{x} = \lim_{t\to 1}\frac{t-1}{t^2-1} =\lim_{t\to 1}\frac{t-1}{(t-1)(t+1)}=\frac12$$

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$x>-1$.

$x= (x +1)- 1= \sqrt{(x+1)}^2-1^2$

$=(\sqrt{(x+1)}-1)(\sqrt{x+1}+1)$.

$\dfrac{\sqrt{(x+1)}-1}{x} =\dfrac{x}{(\sqrt{x+1}+1)x}.$

Take the limit $x \rightarrow 0$.

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