# limit definition of a derivative?

• Dec 5th 2009, 08:47 PM
break
limit definition of a derivative?
consider the following piecewise defined function:

f(x) = {x + 5 if x < 3} and {4*sqrt(x+1) if x >= 3}

Using the limit definition of the derivative, and computing the limits from both sides, show that f(x) is differentiable at x=3.

I have no clue what its asking, can someone help?
• Dec 5th 2009, 08:55 PM
VonNemo19
Quote:

Originally Posted by break
consider the following piecewise defined function:

f(x) = {x = 5 if x < 3} and {4*sqrt(x+1) if x >= 3}

Using the limit definition of the derivative, and computing the limits from both sides, show that f(x) is differentiable at x=3.

I have no clue what its asking, can someone help?

I'm not sure I do either. How can $\displaystyle x=5\text{ if }x<3$?
• Dec 5th 2009, 08:57 PM
break
Quote:

Originally Posted by VonNemo19
I'm not sure I do either. How can $\displaystyle x=5\text{ if }x<3$?

sorry, typo, was supposed to be a + not =
• Dec 5th 2009, 09:16 PM
RockHard
Quote:

Originally Posted by break
sorry, typo, was supposed to be a + not =

Note: this is the same as saying

$\displaystyle \lim_{x\to 3^{-1}} x + 5$
Saying the limit of this function as its approaches values less than 3 to 3, or as the limit as x approaches 3 from the left.

$\displaystyle \lim_{x\to 3^{+1}} 4\cdot \sqrt{x+1}$
Saying the limit of this function as its approaches values more than or equal to 3, or as the limit as x approaches 3 from the right.

Now compute the limit definition of a derivative

$\displaystyle \lim_{h\to 0}\frac{f(x+h)-f(x)}{h^2}$
• Dec 5th 2009, 09:18 PM
VonNemo19
Quote:

Originally Posted by break
sorry, typo, was supposed to be a + not =

You need to show that

$\displaystyle \lim_{x\to3^-}\frac{(x+5)-(3+5)}{x-3}=\lim_{x\to3^+}\frac{4\sqrt{x+1}-4\sqrt{3+1}}{x-3}$
• Dec 5th 2009, 09:33 PM
break
Quote:

Originally Posted by VonNemo19
You need to show that

$\displaystyle \lim_{x\to3^-}\frac{(x+5)-(3+5)}{x-3}=\lim_{x\to3^+}\frac{4\sqrt{x+1}-4\sqrt{3+1}}{x-3}$

uhhh once i simplify it down i get

$\displaystyle \lim_{x\to3^-}\frac{x-3}{x-3}=\lim_{x\to3^+}\frac{4\sqrt{x+1}-8}{x-3}$

do i plug in 3 for the x?
• Dec 5th 2009, 09:40 PM
VonNemo19
Quote:

Originally Posted by break
uhhh once i simplify it down i get

$\displaystyle \lim_{x\to3^-}\frac{x-3}{x-3}=\lim_{x\to3^+}\frac{4\sqrt{x+1}-8}{x-3}$

do i plug in 3 for the x?

Note that the left hand side simplifies to, $\displaystyle \lim_{x\to3^-}1=1$

For the right hand side, multiply both numerator and denominator by the conjugate of the numerator ($\displaystyle \sqrt{x+1}+2$) and see where that gets you.
• Dec 5th 2009, 09:43 PM
break
Quote:

Originally Posted by VonNemo19
Note that the left hand side simplifies to, $\displaystyle \lim_{x\to3^-}1=1$

For the right hand side, multiply both numerator and denominator by the conjugate of the numerator ($\displaystyle \sqrt{x+1}+2$) and see where that gets you.

where did you get ($\displaystyle \sqrt{x+1}+2$) from?
• Dec 5th 2009, 09:57 PM
VonNemo19
Quote:

Originally Posted by break
where did you get ($\displaystyle \sqrt{x+1}+2$) from?

I factored a four out of the numerator.