# limits with some fractions

• Mar 22nd 2013, 08:36 AM
ryu1
limits with some fractions
hi

Attachment 27640

what is the correct way to solve this kind of problem?
do i get an undefined expression with oo * 0 ?
or do i first simplify the right-hand expression and then multiply the two sides and check what happens when x approaches oo ? the limit is oo.
or do i first multiply the two sides and i get the limit to be 11.1?

thanks for the clarification.
• Mar 22nd 2013, 08:54 AM
princeps
Re: limits with some fractions
Quote:

Originally Posted by ryu1
hi

Attachment 27640

what is the correct way to solve this kind of problem?
do i get an undefined expression with oo * 0 ?
or do i first simplify the right-hand expression and then multiply the two sides and check what happens when x approaches oo ? the limit is oo.
or do i first multiply the two sides and i get the limit to be 11.1?

thanks for the clarification.

First simplify the right-hand expression and then multiply the two sides . After that divide both numerator and denominator by $\displaystyle x^2$
• Mar 22nd 2013, 09:01 AM
ryu1
Re: limits with some fractions
Quote:

Originally Posted by princeps
First simplify the right-hand expression and then multiply the two sides . After that divide both numerator and denominator by $\displaystyle x^2$

but if i simplified the hand hand side im left with

(3x-2)(74x+125)

there isnt a denominator here, did i made a mistake?
• Mar 22nd 2013, 09:48 AM
Siron
Re: limits with some fractions
Where's the denominator? Note that
$\displaystyle \frac{6}{4x+7}+\frac{11}{5x+8} = \frac{74x+125}{(4x+7)(5x+8)} = \frac{74x+125}{20x^2+62x+56}$

Can you compute the limit now?
• Mar 22nd 2013, 10:09 AM
ryu1
Re: limits with some fractions
yes then i multiply the numerator by 3x-2 (or just the 74x with the 3x)
I get 222x^2
I divide that by 20x^2
it get me to 11.1 thats the limit?
• Mar 22nd 2013, 02:06 PM
Siron
Re: limits with some fractions
Quote:

Originally Posted by ryu1
yes then i multiply the numerator by 3x-2 (or just the 74x with the 3x)
I get 222x^2
I divide that by 20x^2
it get me to 11.1 thats the limit?

That is correct!
• Mar 22nd 2013, 02:17 PM
ryu1
Re: limits with some fractions
Alright!

I needed a refresher on the simplification of fractions with variables there :)

Thanks!