Question

CALCULUS: is this a correct way to rewrite things
lim x->0 (x^3-7x) / (x^3)
is rewritten as
lim x->0 x^3(1-7x^(-2)) / (x^3)

Answer 1

yup, thats correct.
now u can cancel numerator's x^3 with denominator.
but its not the best way to solve that limit

Answer 2

but the solution says it gives negative infinity

Answer 3

and after rewritten, it gives 1..

Answer 4

NO, after rewritten it does not give 1.
the solution -infinity is correct.
u want to know how to get there ?

Answer 5

YES PLEASE!!

Answer 6

after cancelling x^3 , u get
1-7x^(-2) = 1-7/x^2
right ?
now directly put x=0
what do u get ?

Answer 7

domain error on the calculator..

Answer 8

lol! not on calculator, put it manually...
1-1/0 = ?

Answer 9

does not exist i guess..

Answer 10

ok,
1/0 = infinity
-1/0 = - infinity
now to know whether it exist or has value -infinity, we check for left hand limit and right hand limit.... are u familiar with these terms ?

Answer 11

yes

Answer 12

can u find left hand limit and right hand limit ?

Answer 13

for lim x->0 of 1-7x^2 ?

Answer 14

yes,
whats limit when x-> 0- ?
whats limit when x-> 0+ ?

Answer 15

how do i do that? i just plug the 0? or i plug -0.00000009 for x-> 0- and 0.00000009 for
x->0+

Answer 16

or do i draw a graph?

Answer 17

u plug plug -0.00000009 for x-> 0- and 0.00000009 for
x->0+
but not actually.
just realise that what ever u plug in x^2 in denominator will be positive , isn't it ?

Answer 18

oops, it should be 1-7x^-2

Answer 19

1-7x^(-2) = 1-7/x^2
isn't it ?

Answer 20

so 7/(a number very near to 0)^2 = very large number
1-very large number = -very large number.
whether its 0+ or 0-
so both limit(0+ and 0-) has same value
so lim x->0 exist.
and equals -very large value = -infinity.

Answer 21

thx u are really good!