Calculating Limits for a Quadratic Function

[Dembadon]

Homework Statement

Consider the following function:

$$f(x) = 4x^2 - 8x.$$

Find the limit.

$$\lim_{{\Delta}x\rightarrow 0} \frac{f(x+{\Delta}x)-f(x)}{{\Delta}x}$$

Given: The limit exists.

The Attempt at a Solution

Since the limit exists, I know that I need to do some algebraic manipulations that will enable me to cancel the $${\Delta}x$$ in the denominator.


Here's what I did first:

$$\frac{4(x+{\Delta}x)^2-8(x+{\Delta}x)}{{\Delta}x}$$


After expanding:

$$\frac{4(x^2+2x{\Delta}x+{\Delta}x^2)-8(x+{\Delta}x)}{{\Delta}x}$$


After distributing:

$$\frac{4x^2+8x{\Delta}x+4{\Delta}x^2-8x-8{\Delta}x}{{\Delta}x}$$


Would my next step be?:

$$\frac{(4{\Delta}x^2+8x{\Delta}x-8{\Delta}x)+(4x^2-8x)}{{\Delta}x}$$

...so that I could pull out the $${\Delta}x$$ and cancel it?

 

[Dick]

You forgot to subtract the 'f(x)' that's in the numerator of your difference quotient. That's what cancels the stuff that doesn't have a delta-x in it.

 

[curiousphoton]

Would my next step be?:

$$\frac{(4{\Delta}x^2+8x{\Delta}x-8{\Delta}x)+(4x^2-8x)}{{\Delta}x}$$

...so that I could pull out the $${\Delta}x$$ and cancel it?

The second group (4x^2-8x) in the numerator does not have a {\Delta}x so you can't completely get rid of the {\Delta}x. Do you have any other attemps?

 

[Dembadon]

You forgot to subtract the 'f(x)' that's in the numerator of your difference quotient. That's what cancels the stuff that doesn't have a delta-x in it.

What an embarrassingly careless mistake. Thank you, Dick. I've obtained the correct expression for the limit.

My initial difference quotient should've been:

$$\frac{4(x+{\Delta}x)^2-8(x+{\Delta}x)-(4x^2-8x)}{{\Delta}x}$$

Thanks for the help!