# Thread: Calc: L'Hopital's Rule

1. ## Calc: L'Hopital's Rule

hey can someone please see if I did this right,

x--> infinity, (4x+9) / (7x^2 +4x - 9)
take the derivative = 4 / 14x + 4
so would it be 0? if not i'm not sure what to do next.

and

the integral from 0 to infinity of 1 / (x(x^2)+5)
so i did 1 / x^3 + 5x then i can't take the derivative again b/c the numerator is 1. now i don't know what to do.

any help is much appreciated.

2. Originally Posted by darkblue
hey can someone please see if I did this right,

x--> infinity, (4x+9) / (7x^2 +4x - 9)
take the derivative = 4 / 14x + 4
so would it be 0? if not i'm not sure what to do next..
No need to use it.
Use divide by $x^2$ the numerator and denominator.

3. Originally Posted by darkblue
hey can someone please see if I did this right,

x--> infinity, (4x+9) / (7x^2 +4x - 9)
take the derivative = 4 / 14x + 4
so would it be 0? if not i'm not sure what to do next.
yes, it would be zero.

the integral from 0 to infinity of 1 / (x(x^2)+5)
so i did 1 / x^3 + 5x then i can't take the derivative again b/c the numerator is 1. now i don't know what to do.

any help is much appreciated.
ok. you have two different things here. you said you had 1/(x(x^2) + 5), but if you expand that, you would have 1/(x^3 + 5) not 1/(x^3 + 5x). so i think you made a typo here. please correct it

4. Originally Posted by Jhevon
yes, it would be zero.

ok. you have two different things here. you said you had 1/(x(x^2) + 5), but if you expand that, you would have 1/(x^3 + 5) not 1/(x^3 + 5x). so i think you made a typo here. please correct it
thank you.
oh i see. it is: 1 over x(x^2 + 5), then i just distributed the x.
so 1 over (x^3 + 5x) ?

5. Originally Posted by darkblue
thank you.
oh i see. it is: 1 over x(x^2 + 5), then i just distributed the x.
so 1 over (x^3 + 5x) ?
Don't distribute the x

By the method of partial fractions:
$\frac {1}{x \left( x^2 + 5 \right)} = \frac {A}{x} + \frac {Bx + C}{x^2 + 5}$

$\Rightarrow 1 = (A + B)x^2 + Cx + 5A$

$\Rightarrow A + B = 0$
........... $C = 0$
.......... $5A = 1$

So we have: $A = \frac {1}{5} \mbox { , } B = - \frac {1}{5} \mbox { and } C = 0$

So, $\int \frac {1}{x \left( x^2 + 5 \right)}dx = \int \left( \frac { \frac {1}{5}}{x} - \frac { \frac {1}{5}x}{x^2 + 5} \right)dx$

$= \frac {1}{5} \int \frac {1}{x} dx - \frac {1}{5} \int \frac {x}{x^2 + 5} dx$

Can you take it from here?

6. ohh Partial Fractions! I forgot about that. thankyou

so i got (1/5 ln x - (1/10) ln(x^2 + 5) + C
is the 1/10 right? because i multiplied 1/5 by 1/2.

7. Originally Posted by darkblue
ohh Partial Fractions! I forgot about that. thankyou

so i got (1/5 ln x - (1/10) ln(x^2 + 5) + C
is the 1/10 right? because i multiplied 1/5 by 1/2.
yes, that's correct

8. thanx again! problem solved!