First Principles (2) (1 Viewer)

atar90plus · Feb 6, 2012

Sorry, I decided to post this question again because I typed wrong before. Also could you guys please help me with these 2 limits questions. I tried doing them but my answer was wrong. Could you guys please show full working out and solutions.

14. f(x) = x+2/x-3 at x=1

4. lim
x->root 3 root 3x - x / 3 -x^2

5. lim
x->root a x^2 - a / x+ root a

Thanks

IamBread · Feb 6, 2012

atar90plus said:
Sorry, I decided to post this question again because I typed wrong before. Could you guys please show full working out and solutions

14. f(x) = x+2/x-3 at x=1

Thanks

$f(x) = \frac{x+2}{x-3}$

$f(x) = \frac{x+5-3}{x-3}$

$f(x) = 1 + \frac{5}{x-3}$

$f(x+h) = 1 + \frac{5}{x+h-3}$

$f'(x) = \lim_{h \to 0} \frac{1 +\frac{5}{x+h-3} - (1 + \frac{5}{x-3})}{h}$

$= \lim_{h \to 0} \frac{\frac{5}{x+h-3} - \frac{5}{x-3}}{h}$

$=\lim_{h \to 0}5\frac{\frac{x-3 - x -h +3}{(x+h-3)(x-3)}}{h}$

$=\lim_{h \to 0}5\frac{\frac{-h}{(x+h-3)(x-3)}}{h}$

$=\lim_{h \to 0}-5\frac{1}{(x+h-3)(x-3)}$

$=\lim_{h \to 0}-5\frac{1}{x^2 -6x+xh+9-3h}$

$f'(x) = -\frac{5}{x^2 -6x +9}$

$f'(1) = -\frac{5}{1-6+9}$

$f'(1) = -\frac{5}{4}$

Shadowdude · Feb 6, 2012

atar90plus said:
Sorry, I decided to post this question again because I typed wrong before. Could you guys please show full working out and solutions

14. f(x) = x+2/x-3 at x=1

Thanks

$\\\textup{Find\,}f'(1)\,\textup{of}\;f(x)=\frac{x+2}{x-3}$
$\textup{We note:}\;\frac{x+2}{x-3} = \frac{x-3+5}{x-3}=\frac{x-3}{x-3}+\frac{5}{x-3}=1+\frac{5}{x-3}$

So then:

$\\ \frac{d}{dx}\frac{x+2}{x-3}=\frac{d}{dx}\left (1 + \frac{5}{x-3} \right )$

$=\lim_{h \to 0} \frac{\left (1 + \frac{5}{x+h-3} \right ) - \left (1 + \frac{5}{x-3} \right )}{h}$

Now, the 1s cancel:

$=\lim_{h \to 0} \frac{\frac{5}{x+h-3}-\frac{5}{x-3}}{h}$

Take the 5 out the front:

$=5\lim_{h \to 0} \frac{\frac{1}{x+h-3}-\frac{1}{x-3}}{h}$

Cross-multiply:

$=5\lim_{h \to 0} \frac{\frac{(x-3)-(x+h-3)}{(x-3)(x+h-3)}}{h}$

Simplify the top:

$=5\lim_{h \to 0} \frac{\frac{-h}{(x-3)(x+h-3)}}{h}$

Take the negative out the front:

$=-5\lim_{h \to 0} \frac{h}{h(x-3)(x+h-3)}$

Re-write the expression to make it nice, and cancel the h's:

$=-5\lim_{h \to 0} \frac{1}{(x-3)(x+h-3)}$

Take the limit and simplify:

$=-5 \frac{1}{(x-3)(x-3)} = -\frac{5}{(x-3)^{2}}$

So:

$f'(x) = -\frac{5}{(x-3)^{2}}$

Then:

$f'(1) = -\frac{5}{(1-3)^{2}} = -\frac{5}{(-2)^{2}} = -\frac{5}{4}$

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IAmBread, you have an error here:

IamBread said:
$=\lim_{h \to 0}5\frac{\frac{x-3 - x -h +3}{(x+h-3)(x-3)}}{h}$

$=\lim_{h \to 0}5\frac{\frac{h}{(x+h-3)(x-3)}}{h}$

It should be -h, not +h.

IamBread · Feb 6, 2012

Shadowdude said:
---

IAmBread, you have an error here:

It should be -h, not +h.

Ah yes, thanks. Always forget the minus's... and I forgot to sub in 1

ah well. I'll fix it.

First Principles (2) (1 Viewer)

atar90plus

01000101=YES! YES! YES!

IamBread

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Shadowdude

Cult of Personality

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