Thread: L'Hospitals rule and Mean Value Theorem

1. L'Hospitals rule and Mean Value Theorem

1) $\displaystyle F(x) = \frac{f(1+5x)+f(1+6x)}{x}$

Given f(1) = 0, and f '(1) = 8, evaluate $\displaystyle \lim_{x\to0}F(x)$

Limits as it results to $\displaystyle \frac{0}{0}$, so differentiate top and bottom using Chain Rule:

$\displaystyle \lim_{x\to0} \frac{(5)f'(1+5x)+(6)f'(1+6x)}{1} = 40+48 = 98$

2) If f(2) = 13 and f '(x) ≥ 1 for 4 ≤ x ≤ 4, how small can f(4) possibly be?

f(4) - f(2) = f'(c)(4-2)
f(4) = 13 + (2)f'(c) > 1

?

The last step, and where to go from there is what I'm a little shaky on.

3) $\displaystyle \lim_{x\to\infty} (1+\frac{c}{x})^q$
where q = hx
h and c are both constants, to my knowledge.

We were taught to do this in a way that seems a little convoluted, but here's as far as I got with this one:

$\displaystyle \lim_{x\to\infty} (1+\frac{c}{x})^q$ = $\displaystyle (e^z)^q = e^g$

$\displaystyle z = ln(1+\frac{c}{x}); g = (hx)ln(1+\frac{c}{x})$

From that point we can just take the limit of e to the g'(x), so

$\displaystyle g(x) = (hx)ln(1+\frac{c}{x})$ = $\displaystyle \frac{ln(1+\frac{c}{x})}{\frac{1}{hx}}$

Then I get stuck again, because the derivative of the bottom is $\displaystyle \frac{-h}{(hx)^2}$

I'm imaging there has to be a way to simplify that, but I can't think of it.

2. Originally Posted by Open that Hampster!
1) $\displaystyle F(x) = \frac{f(1+5x)+f(1+6x)}{x}$

Given f(1) = 0, and f '(1) = 8, evaluate $\displaystyle \lim_{x\to0}F(x)$

Limits as it results to $\displaystyle \frac{0}{0}$, so differentiate top and bottom using Chain Rule:

$\displaystyle \lim_{x\to0} \frac{(5)f'(1+5x)+(6)f'(1+6x)}{1} = 40+48 = 98$

$\displaystyle {\color{red}\mbox{ Of course, you're assuming here that } f'(x) \mbox{ is continuous at } x = 1,}$ $\displaystyle {\color{red}\mbox{ otherwise this is unjustified}}$

2) If f(2) = 13 and f '(x) ≥ 1 for 4 ≤ x ≤ 4, how small can f(4) possibly be?

f(4) - f(2) = f'(c)(4-2)
f(4) = 13 + (2)f'(c) > 1

?

The last step, and where to go from there is what I'm a little shaky on.

$\displaystyle {\color{blue}\mbox{ You're almost there:} f(4)=13+2f'(c)>13+2=15}$

3) $\displaystyle \lim_{x\to\infty} (1+\frac{c}{x})^q$
where q = hx
h and c are both constants, to my knowledge.

We were taught to do this in a way that seems a little convoluted, but here's as far as I got with this one:

$\displaystyle \lim_{x\to\infty} (1+\frac{c}{x})^q$ = $\displaystyle (e^z)^q = e^g$

$\displaystyle {\color{red}\mbox{ I don't have the faintest idea what you did here. Imo it is way simpler:}}$
$\displaystyle {\color{red}\lim_{x\rightarrow\infty}\left(1+\frac {c}{x}\right)^q=\left(\lim_{x\rightarrow\infty}\le ft(1+\frac{c}{x}\right)^x\right)^h=e^{ch}}$

$\displaystyle {\color{blue}Tonio}$

$\displaystyle z = ln(1+\frac{c}{x}); g = (hx)ln(1+\frac{c}{x})$

From that point we can just take the limit of e to the g'(x), so

$\displaystyle g(x) = (hx)ln(1+\frac{c}{x})$ = $\displaystyle \frac{ln(1+\frac{c}{x})}{\frac{1}{hx}}$

Then I get stuck again, because the derivative of the bottom is $\displaystyle \frac{-h}{(hx)^2}$

I'm imaging there has to be a way to simplify that, but I can't think of it.
.