Find the Limit of f(x) = x^2 as x -> a

[Jade]

f(x) = x^2

limit x goes to a

[f(a + h) - f(a)] / h

does this problem require finding the slope of a tangent line

 

[galactus]

Yes, it does. This is the classic definition of a derivative.

Use the limit to find the derivative of x^2.

You probably know that the derivative of x^2 is 2x, but can you use the

limit to find it?.

 

[Jade]

hmmmm

:?

This is what is going through my mind right now

Derivative is the slope of the tangent line

the derivative of x^2 is 2x

f(x)=mx+b

f( )=m( ) +b

I think I am making this more difficult than it needs to be

 

[galactus]

Yes, you're kind of on the right track. Just as in a line equation(point-slope form), \(\displaystyle slope=\frac{y_{1}-y}{x_{1}-x}\)

y=f(x)

So, you have:

\(\displaystyle \L\\m=\lim_{x_{1}\rightarrow{x}}\frac{f(x_{1})-f(x)}{x_{1}-x}\)

We rewrite this formula in terms of h as \(\displaystyle h=x_{1}-x\)

Thus, \(\displaystyle x_{1}=x+h\) and \(\displaystyle h\rightarrow{0}\) as \(\displaystyle x_{1}\rightarrow{x}\)

Therefore, we can rewrite as

\(\displaystyle m=\L\\\lim_{h\to\0}\frac{f(x+h)-f(x)}{h}\)

Using your function \(\displaystyle f(x)=x^{2}\)

\(\displaystyle \L\\\lim_{h\to\0}\frac{(x+h)^{2}+x^{2}}{h}\)

Now, can you take the limit and find the derivative?.

 

[Jade]

Derivative

:?

derivative of x is 1

derivative of a contant is 0

x^2 is 2x which is 2

using f(x)=x^2

my brain is wanting me to say f(1)=2

(1+0)^2+2/0

This is not right ?

 

[galactus]

No, take the limit.

Expand: \(\displaystyle \L\\\lim_{h\to\0}\frac{(x+h)^{2}-x^{2}}{h}\)

=\(\displaystyle \L\\\lim_{h\to\0}\frac{x^{2}+2xh+h^{2}-x^{2}}{h}\)

=\(\displaystyle \L\\\lim_{h\to\0}\frac{2xh+h^{2}}{h}\)

=\(\displaystyle \L\\\lim_{h\to\0}\frac{2x\sout{h}}{\sout{h}}+\lim_{h\to\0}\frac{h^{\sout{2}}^{1}}{\sout{h}}\)

=2x

See?. That's your derivative. That's what a derivative is.

This is the core idea behind a derivative.

 

[Jade]

You are so smart

Thank you so much