Sqrt (x) derivative by first principles

[apple2357]

I have seen a proof of the derivative of sqrt(x) by first principles and one approach seems to go a little like this:



My question concerns the penultimate line? We can't actually substitute h=0 into the expression, how can we be sure the limit evaluated is correct?

 

[willyengland]

We can't actually substitute h=0 into the expression,

Why not? We are not dividing by zero or anything.

 

[apple2357]

No, but i thought when we take a limit we never actually substitute h=0, we let h tend towards zero which is subtly different?

 

[willyengland]

Strictly speaking you should not write:
[MATH] \lim_{h \to 0} \frac{1}{\sqrt{x+h}+\sqrt{x} } = \frac{1}{\sqrt{x+0}+\sqrt{x} } [/MATH]
but only:
[MATH] \lim_{h \to 0} \frac{1}{\sqrt{x+h}+\sqrt{x} } = \frac{1}{2\sqrt{x}} [/MATH]

 

[apple2357]

Yes but then it feels like there is a missing step in that last line of work ( or missing explanation?). How do we actually know it is true?

 

[Harry_the_cat]

Here's a question:
What is lim (x->0) 2x +3 ?

 

[apple2357]

Here's a question:
What is lim (x->0) 2x +3 ?

This feels much simpler, I recognise it and know that there is nothing unusual about it, it's linear etc. But thats not the case for the expression involving reciprocal square roots above.

 

[HallsofIvy]

No, but i thought when we take a limit we never actually substitute h=0, we let h tend towards zero which is subtly different?

You thought wrong. For any function that is continuous at x= a, \(\displaystyle \lim_{x\to a} f(x)= f(a)\).
(That's pretty much the definition of "continuous".) And as long as we have no square roots of negative numbers or divisions by 0 a function like this is continuous. It's only when we have square roots of negative numbers or divisions by 0 that we have to be "subtle".

 

[apple2357]

Thanks! Can you give me an example where a limit exists but substituting would be wrong?

 

[Dr.Peterson]

How about any place earlier than the penultimate line in the proof you are asking about?

As you were told, if a function is continuous, you can substitute; you can't substitute when it is not continuous, and in particular when it isn't even defined there.

In many instances of working out a limit, the process amounts to rewriting and simplifying until you get a new function that is equal to the original everywhere except the place where it previously was undefined, but now is defined and continuous there. Then you can substitute.

 

[apple2357]

Thanks all for clearing that up for me.

 

[HallsofIvy]

Thanks! Can you give me an example where a limit exists but substituting would be wrong?

Let f(x) be defined as "f(x)= 3x if x is not equal to 2 and f(2)= 0". Substituting" 2 for x gives 0 but the limit is 6.

Or let f(x)= sin(x)/x. Substituting 0 for x gives the "indeterminate" 0/0 but the limit is 1.

Or let f(x)= 0 if x< 1, f(x)= 5 if $x\ge 2$. Substituting 1 for x gives 5 but there is no limit.