# Thread: Need help on L'Hospitals, Partial fractions

1. ## 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:
$\displaystyle \int \frac{4y^2-7y-12}{y(y+2)(y-3)}dy$

2. #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.

3. Hello, c_323_h!

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

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

5. 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,
$\displaystyle \lim_{x\to\pi/2^+}\frac{-\sin x}{-\cos x}=\lim_{x\to\pi/2^+}\tan x=+\infty$
---
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 $\displaystyle x\to \pi/2^+$