# Need help on L'Hospitals, Partial fractions

• Jul 25th 2006, 01:44 PM
c_323_h
Need help on L'Hospitals, Partial fractions
Hi guys I have a test tomorrow and am confused on a few homework questions on L'Hospital's Rule and partial fractions

1. Using L'Hospital's Rule:
$\displaystyle \lim_{x\rightarrow{0}}\frac{e^x-1-x}{x^2}$

2. L'Hospital's Rule:
$\displaystyle \lim_{x\rightarrow{{\pi/2}^+}}\frac{cosx}{1-sinx}$

3. Partial Fractions:
$\displaystyle \int \frac{4y^2-7y-12}{y(y+2)(y-3)}dy$
• Jul 25th 2006, 04:04 PM
galactus
#1.

Do L'hopital twice.

That is, take the derivative of the numerator and denominators twice

$\displaystyle \frac{d}{dx}[e^{x}-x-1]=e^{x}-1$

$\displaystyle \frac{d}{dx}[e^{x}-1]=e^{x}$

The second derivative of $\displaystyle x^{2}$ is 2.

You have: $\displaystyle \lim_{x\to\0}\frac{e^{x}}{2}$

Now, as x approaches 0, you have 1/2

For #2. I believe you may have a 'tan' in your future.

For #3. $\displaystyle \frac{A}{y}+\frac{B}{y+2}+\frac{C}{y-3}=4y^{2}-7y-12$

$\displaystyle A(y+2)(y-3)+B(y-3)y+C(y+2)y=4y^{2}-7y-12$

$\displaystyle Ay^{2}+By^{2}+Cy^{2}-Ay-3By+2Cy-6A=4y^{2}-7y-12$

Equate coefficients:

$\displaystyle \left[\begin{array}{ccc|c}A&B&C&4\\-A&-3B&2C&-7\\-6A&0&0&-12\end{array}\right]$

Solve the system.

Soroban has a slick way of doing partial fractions. I trust he'll be along to demonstrate.
• Jul 25th 2006, 07:57 PM
Soroban
Hello, c_323_h!

Cody (Galactus) is right . . . there is a slick way.
You may already know it, though.

Quote:

3. Partial Fractions: .$\displaystyle \int \frac{4y^2-7y-12}{y(y+2)(y-3)}dy$

We have the set up: .$\displaystyle \frac{4y^2 - 7y - 12}{y(y+2)(y-3)}\:=\:\frac{A}{y} + \frac{B}{y+2} + \frac{C}{y-3}$

Multiply through by the LCD:
. . $\displaystyle 4y^2 - 7y - 12 \:= \:A(y+2)(y-3) + By(y-3) + C(y(y-2)$

Now pick some "good" values for $\displaystyle y$ *

Let $\displaystyle y = 0:\;\;-12\:=\:A(2)(-3) + B(0) + C(0)\quad\Rightarrow\quad-6A =$$\displaystyle -12\quad\Rightarrow\quad \boxed{A = 2} Let \displaystyle y = -2:\;\;18 \:=\:A(0) + B(-2)(-5) + C(0)\quad\Rightarrow\quad10B =$$\displaystyle 18\quad\Rightarrow\quad \boxed{B = \frac{9}{5}}$
Let $\displaystyle y = 3:\;\;3 \:=\:A(0) + B(0) + C(3)(5)\quad\Rightarrow\quad 15C =$$\displaystyle 3\quad\Rightarrow\quad \boxed{C = \frac{1}{5}} Therefore: .\displaystyle \frac{4y^2 - 7y - 12}{y(y+2)(y-3)} \:=\:\frac{2}{y} + \frac{\frac{9}{5}}{y + 2} + \frac{\frac{1}{5}}{y-3} ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ * I hope it's obvious that "good" values are . . those that create zeros in the expression. • Jul 25th 2006, 08:35 PM c_323_h Quote: Originally Posted by Soroban Hello, c_323_h! Cody (Galactus) is right . . . there is a slick way. You may already know it, though. We have the set up: .\displaystyle \frac{4y^2 - 7y - 12}{y(y+2)(y-3)}\:=\:\frac{A}{y} + \frac{B}{y+2} + \frac{C}{y-3} Multiply through by the LCD: . . \displaystyle 4y^2 - 7y - 12 \:= \:A(y+2)(y-3) + By(y-3) + C(y(y-2) Now pick some "good" values for \displaystyle y * Let \displaystyle y = 0:\;\;-12\:=\:A(2)(-3) + B(0) + C(0)\quad\Rightarrow\quad-6A =$$\displaystyle -12\quad\Rightarrow\quad \boxed{A = 2}$

Let $\displaystyle y = -2:\;\;18 \:=\:A(0) + B(-2)(-5) + C(0)\quad\Rightarrow\quad10B =$$\displaystyle 18\quad\Rightarrow\quad \boxed{B = \frac{9}{5}} Let \displaystyle y = 3:\;\;3 \:=\:A(0) + B(0) + C(3)(5)\quad\Rightarrow\quad 15C =$$\displaystyle 3\quad\Rightarrow\quad \boxed{C = \frac{1}{5}}$

Therefore: .$\displaystyle \frac{4y^2 - 7y - 12}{y(y+2)(y-3)} \:=\:\frac{2}{y} + \frac{\frac{9}{5}}{y + 2} + \frac{\frac{1}{5}}{y-3}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

* I hope it's obvious that "good" values are
. . those that create zeros in the expression.

yeah...this is the way they taught me in class. just couldn't quite the decomposition right. thanks

thanks to galactus too
• Jul 25th 2006, 08:39 PM
ThePerfectHacker
Quote:

Originally Posted by c_323_h
2. L'Hospital's Rule:
$\displaystyle \lim_{x\rightarrow{{\pi/2}^+}}\frac{cosx}{1-sinx}$

They give, $\displaystyle \frac{0}{0}$ use L'HopitAl
Derivative of numerator,
$\displaystyle -\sin x$
Derivative of denominator,
$\displaystyle -\cos x$
Thus,
$\displaystyle \lim_{x\to\pi/2^+}\frac{-\sin x}{-\cos x}=\lim_{x\to\pi/2^+}\tan x=+\infty$
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I recommend to avoid L'Hopital but rather do,
$\displaystyle \frac{\cos x}{1-\sin x}\cdot \frac{1+\sin x}{1+\sin x}=\frac{\cos x(1+\sin x)}{1-\sin^2x}=$$\displaystyle \frac{\cos x(1+\sin x)}{\cos ^2x}=\frac{1+\sin x}{\cos x}=\sec x+\tan x\to \infty$ as $\displaystyle x\to \pi/2^+$