Limit

Derek

Evaluate the limit as h approaches 0:

$\frac {1}{h} (\frac {1}{\sqrt {1+h}}-1)$

**Posted:** Tue Sep 20, 2005 1:47 pm

riddler

what is a LIMIT?

**Posted:** Tue Sep 20, 2005 2:02 pm

chess64

have you taken (pre)calculus? He wants you to find $\lim_{h\rightarrow 0} \frac{1}{h} \left( \frac{1}{\sqrt{1+h}}-1 \right)$ . If you don't know what that is, it's just the value that $f(h)$ gets closer and closer to...

riddler

**Posted:** Tue Sep 20, 2005 2:10 pm

jianuovidiu

**Posted:** Tue Sep 20, 2005 4:30 pm

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Kent Merryfield

Let's treat this as an exercise somewhere near the boundary between precalculus and first-semester calculus. In other words, we never heard of L'Hôpital, and maybe don't yet know what a derivative is.

$\lim_{h\to0}\frac1h\left(\frac1{\sqrt{1+h}}-1\right).$

Let's see - if $h$ is very small, then $\sqrt{1+h},$ and its reciprocal, must be very near 1. Subtract 1 and you get something very small. Now divide by $h$ - uh-oh, what's a very small number divided by a very small number? The answer to that is "It depends - how small and how small?" We can't settle the issue without doing some more work. (By the way, that's what " $\frac00$ " means: not literally zero over literally zero, but very small over very small.)

Since we don't know any calculus yet, we must fall back on what we do know: algebra.

$\frac1h\left(\frac1{\sqrt{1+h}}-1\right)= \frac1h\left(\frac{1-\sqrt{1+h}}{\sqrt{1+h}}\right)$

At this point, note that the $\sqrt{1+h}$ in the denominator is harmless - it might as well be 1. We don't need to do anything about it. We do need to do something about the $1-\sqrt{1+h}$ in the numerator, which is far from harmless, being a non-obvious version of "very small." So we attack that - we choose to rationalize the numerator.

$\frac1h\left(\frac{1-\sqrt{1+h}}{\sqrt{1+h}}\right)= \frac1h\left(\frac{1-\sqrt{1+h}}{\sqrt{1+h}}\cdot\frac{1+\sqrt{1+h}} {1+...$

$=\frac{-h}{h\sqrt{1+h}\,(1+\sqrt{1+h})}= \frac{-1}{\sqrt{1+h}\,(1+\sqrt{1+h})}$

Now let $h$ tend to zero. All the parts of the denominator are now harmless and predictable, and we can see that the limit must be $-\frac12.$

What in that solution was the least obvious and required the most insight? The choice to rationalize the numerator rather than the denominator. Can you see why I did that?

**Posted:** Tue Sep 20, 2005 5:03 pm

Kent Merryfield

Of course, if you free me from the "boundary of precalculus and Calculus I" constraint and let me use all of my favorite tools, I'll use Newton's Binomial Theorem for this one:

$\lim_{h\to0}\frac1h\left((1+h)^{-\frac12}-1\right)= \lim_{h\to0}\frac1h\left((1-\frac h2+O(h^2))-1\right)$

$=\lim_{h\to0}\left(-\frac12+O(h)\right)=-\frac12.$

**Posted:** Tue Sep 20, 2005 5:09 pm

blahblahblah

Besides, this is just the derivative of $(1+x)^{-\frac 12}$ at $x=0$ ; using L'Hopital would be circular.

**Posted:** Tue Sep 20, 2005 5:16 pm

Derek

Yes we assume that the person has never heard of L'Hopital rule before, although it would be convenient to use it.

Thanks guys.

**Posted:** Tue Sep 20, 2005 6:41 pm