Question

How do i solve a limit when x approaches infinity?????

Answer 1

There is no general method
for most of them
we can do a substitution x=1/y
and take y approaching 0

do u have any specific question ?

Answer 2

this website would help:
http://www.mathsisfun.com/calculus/limits-infinity.html

Answer 3

yes....6x^2+5x+4/2x^2-1

Answer 4

then you can plug in y=1/x as i told

but there's an easier way out too

divide numerator and denominator by highest power of x
here x^2

Answer 5

take coefficent of highest power of numerator and denominator

Answer 6

Thanks.... so is the correct answer -2?

Answer 7

no, how did u get -2 ?

Answer 8

mmm whats the coefficent of highest power of numerator ?

Answer 9

by dividing the numerator and the denominator by x^2 but i must have got lost somewhere.:):)

Answer 10

\(\Large \dfrac{6x^2+5x+4}{2x^2-1}\)

\(\Large \dfrac{6x^2/x^2+5x/x^2+4/x^2}{2x^2/x^2-1/x^2}\)

now
x/x^2 =1/x

when x approached infinity
1/x and 1/x^2 approaches 0

so,
\(\Large \dfrac{6+0+0}{2-0}\)

see if you get this

Answer 11

when highest exponents of the variable are same (here 2),
then you can mentally calculate the value of limit

just take the co-efficient of highest powers of x in numerator and deominator as ikram said :)

so, for numerator, its 6
denominator its 2
so limit =6/2
:)

Answer 12

ok here is a Hint :-
if n>m then :-

\(\large lim_n \to \infty \left ( \frac{a_0x^n+a_1x^{n-1}+a_1x^{n-2}+.....+a_nx^{n-n} }{b_0x^m+b_1x^{m-1}+b_1x^{m-2}+.....+b_mx^{m-m} } \right )= \frac{a_0}{b_0} \)

Answer 13

that applies only when
n=m

Answer 14

for n>m
Limit is infinity

for n<m
limit =0

Answer 15

wait sorry xD yeah typo

Answer 16

My equations dont seem to be working...........they are still coming as "\(\large lim_n \to \infty \left ( \frac{a_0x^n+a_1x^{n-1}+a_1x^{n-2}+.....+a_nx^{n-n}" etc

Answer 17

refresh your page once...

Answer 18

http://prntscr.com/4afu4f

Answer 19

ok this time better :3

Answer 20

Yeah ok............i'm understanding it now........thanks all

Answer 21

cool :)
ask if anymore doubts!

Answer 22

thanks again