Help with Calculus problem? It's pretty weird.?

Question

Ok it asks to Compute:

Delta f(x) = lim as h -> 0 of (f(x+h) + f(x-h) - 2f(x))/h^2 
For the following:
a.) f(x) = x
b.) f(x) = x^2
and c.) f(x) = x^3

And also, it says "based on these examples, what do you think the limit Delta f(x) represents?

This is what I got. (So you don't think I'm just trying to get the answer)

Ok, I thought this looks eerily similar to the definition of the derivative, so I plugged in x and got (x+h+x-h-2x)/h^2. When you plug in 0 for h and cancel the like terms, you get 0/0. So I thought ok ill just do the next one. And I got 0/0 again... :/ and guess what? For c.) I got 0/0 too...

Is it that easy or am I missing something? And I believe that delta f(x) represents possibly that the change in f(x) is indeterminable/undefined as h approaches zero for any function. Does that sound right?

teekshi33 · Accepted Answer

Hello! You are kind of on the right track, but not really, haha.

So, when f(x) = x, you are right that Δf(x) = lim as h -> 0 of [(x + h + x − h − 2x) / h²]

Now, don't plug in 0 yet, but first cancel the like terms: Δf(x) = lim as h -> 0 of [0 / h²]

Plugging in 0 does yield 0/0, but this is an indeterminate form. Have you learned L'Hospital's Rule yet? Basically, it says that if taking a limit of something divided by something else gives an indeterminate form (like 0/0 or ∞/∞), then you can set this equal to the limit of the derivative of the top divided by the derivative of the bottom and continue solving. So this becomes:

Δf(x) = lim as h -> 0 of [0 / 2h] ... This is still 0/0 so we do L'Hospital's again:
= lim as h -> 0 of [0 / 2] 
= 0

Now we do it for f(x) = x². Be really careful here! You are NOT just plugging in x² to what we did above. 

Δf(x) = lim as h -> 0 of [((x + h)² + (x − h)² − 2x²) / h²]
= lim as h -> 0 of [(x² + 2xh + h² + x² −  2xh + h² − 2x²) / h²]
= lim as h -> 0 of [2h² / h²]
= lim as h -> 0 of [2]
= 2

Finally, we do it for f(x) = x³. Signs are really tricky here:

Δf(x) = lim as h -> 0 of [((x + h)³ + (x − h)³ − 2x³) / h²]
= lim as h -> 0 of [(x³ + 3x²h + 3xh² + h³ + x³ − 3x²h + 3xh² + h³ − 2x³) / h²]
= lim as h -> 0 of [(6xh² + 2h³) / h²]
= lim as h -> 0 of [h²(6x + 2h) / h²]
= lim as h -> 0 of [6x + 2h]
= 6x

So we have (a) 0, (b) 2, (c) 6x. What could this mean? It turns out that Δf(x) actually represents the second derivative of f(x), or the derivative of the derivative. Try it out:

f(x) = x.... f'(x) = 1.... f''(x) = 0
f(x) = x².... f'(x) = 2x.... f''(x) = 2
f(x) = x³.... f'(x) = 3x².... f''(x) = 6x

Cool, right?? I don't know how you would be expected to know this just from examining the answers, though. haha.

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Help with Calculus problem? It's pretty weird.?

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