Calculus Problem of the Week September 30, 2011

Sometimes easy and sometimes hard, our calculus problem of the week could come from any calculus topic. If you really want to get better at calculus, following these problems is a great way to make yourself practice! Past calculus problems of the week.

This week’s problem:
(click “see the solution” at the bottom of post to, well, see the solution. If it doesn’t work, please click the title of this post and then try (I’m working on this!))

Find the limit below and explain what it represents.
\lim_{h\to 0}\dfrac{\sqrt{x+1+h}-\sqrt{x+1}}{h}.
.

See the solution.

This limit is actually f'(x) for f(x)=\sqrt{x+1}. Recall the definition of the derivative of a function at a point x:

\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}. (sometimes delta X is used instead of h)
If you simply try to plug in zero for h, you will get 0/0 which is an indeterminate form. You can think of this as basically giving you now information other than “try another way!”.
One of the “classic” techniques to deal with this is by rationalizing the numerator (or denominator if that is where the “root functions” are) by multiplying it by the conjugate. The conjugate will have the exact same terms but a different sign in the middle. In this instance, the conjugate is \sqrt{x+1+h}+\sqrt{x+1}. Two things to keep in mind:

  1. Anytime you perform an operation like this to a fraction, you must perform it to both the numerator and the denominator. That way, it still has the same value, it just looks different.
  2. Since the conjugate has an opposite sign, you will eliminate all of the terms with roots – which is the whole reason to use it! BUT, you will have these terms appear where they weren’t before. Usually this is OK.

So, multiplying by the conjugate in both the numerator and denominator in this example:
\lim_{h\to 0}\dfrac{\sqrt{x+1+h}-\sqrt{x+1}}{h} = \lim_{h\to 0}\dfrac{\sqrt{x+1+h}-\sqrt{x+1}}{h} \left(\dfrac{\sqrt{x+1+h}+\sqrt{x+1}}{\sqrt{x+1+h}+\sqrt{x+1}}\right)
Now, cancel out the h’s and then plug in zero for any remaining h:
\lim_{h\to 0}\dfrac{(x+h+1)-(x+1)}{h(\sqrt{x+1+h}+\sqrt{x+1})}=\lim_{h\to 0}\dfrac{h}{h(\sqrt{x+1+h}+\sqrt{x+1})}=\lim_{h\to 0}\dfrac{1}{\sqrt{x+1+h}+\sqrt{x+1}}=\dfrac{1}{2\sqrt{x+1}}.

Rationalizing the numerator is what allowed me to find this limit “simply” by plugging in the zero. This technique should be one you remember, as it comes up pretty regularly!