Question

Can someone help me understand this problem?
find the limit of (F(x+deltax)-f(x))/deltax as deltax--->0, and f(x) x^2+2
When i did it, i got 1, because the f(x)-f(x) canceled each other out in the numerator and it'd be deltax/delta x which is always 1 except at 0, so the limit would be 1. Is that correct?

Answer 1

I would agree with you

Answer 2

\[\lim_{\Delta x~\rightarrow ~0} \dfrac{f(x+\Delta x)-f(x)}{\Delta x} \\~\\~\\ = \lim_{\Delta x~\rightarrow ~0}\dfrac{[(x+\Delta x)^2+2] - [x^2+2]}{\Delta x}\\~\\~\\ = \lim_{\Delta x~\rightarrow ~0}\dfrac{[x^2+2x\Delta x+\Delta x^2+2] -[x^2+2]}{\Delta x}\\~\\~\\ = \lim_{\Delta x~\rightarrow ~0}\dfrac{[\cancel{x^2}+2x\Delta x+\Delta x^2+\cancel{2}] -[\cancel{x^2}+\cancel{2}]}{\Delta x}\\~\\~\\ = \lim_{\Delta x~\rightarrow ~0}\dfrac{2x\Delta x+\Delta x^2}{\Delta x}\\~\\~\\ = \lim_{\Delta x~\rightarrow ~0}\dfrac{\Delta x(2x+\Delta x)}{\Delta x}\\~\\~\\ = \lim_{\Delta x~\rightarrow ~0}\dfrac{\cancel{\Delta x}(2x+\Delta x)}{\cancel{\Delta x}}\\~\\~\\ = \lim_{\Delta x~\rightarrow ~0}2x+\Delta x \\~\\= 2x+0=\boxed{2x}\]

Answer 3

So by simplification, you would get 2x, instead of 1.

Does my work make sense?

Answer 4

I guess? So you cant cancel stuff out like that(the way i explained in my question)? It doesnt work that way? Sorry i didnt respond earlier was busy

Answer 5

None, it don't work that way.

The way you are thinking is like \[\large \displaystyle \lim_{\Delta x~\rightarrow~0}\dfrac{f(x+\Delta x)-f(x)}{\Delta x}\Longrightarrow\dfrac{\lim_{\Delta x~\rightarrow~0 f(x+\Delta x)-f(x)}}{\lim_{\Delta x~\rightarrow~0} \Delta x}\]

Then you will have 0 in denominator (by direct substitution), which is a big no-no.

Does that make sense?

Answer 6

Yea, got it