Question

lim x->infinity ((2x+5)(x-2))/((7x-2)(3x+1))

Answer 1

divide numerator and denominator by x^2

Answer 2

so that you get the terms like 1/x , 1/x^2
since x ->infinity,
1/x -> 0 , 1/x^2 ->0

Answer 3

Since the numerator's highest term is 2x^2, and the denominator's highest term is 21x^2, the limit will be 2/21, the ratio of the coefficients of x^2.

Answer 4

did you try dividing x^2?
whats your doubt ?

Answer 5

I am kind of stuck on that part.

Answer 6

You could divide each term in the numerator and denominator by x^2, and take lim x---> infinity for each term, and you'll get 2/21. Thats a fine approach.

I was just showing a different approach, which is a consequence of dividing each term by x^2.

Answer 7

what do you get if you divide 2x+5 by x ??

Answer 8

would you get this :
2 +5/x
?

Answer 9

2+5/x

Answer 10

Sorry it took me a while.

Answer 11

no problem,
so
(x-2 )/x =...
(7x-2 )/x = ...
(3x+1) /x =....

Answer 12

(x-2 )/x =1+ -2/x
(7x-2 )/x = 7+-2/x
(3x+1) /x = 3+-1/x

Answer 13

last one,
3 +1/x

Answer 14

so, now since x ->infinity
1/x -> 0
so, you can just put 0 wherever you see the denominator as x

Answer 15

So does that need to be solved? After I plug in wherever the 0 is?

Answer 16

Also is it undefined like 5/0?

Answer 17

no...

((2x+5)(x-2))/((7x-2)(3x+1))
= (2+5/x) (1-2/x) / (7-2/x)( 3+1/x)

so put 5/x = 0
2/x=0
1/x =0
in that, what remains ?

Answer 18

Oh I see. Its 2/21.

Answer 19

good, thats correct :)

Answer 20

Thank you so much @hartnn and @Easyaspi314.

Answer 21

welcome ^_^