Question

Calculus Limits at Infinity
Please help?

find lim h(x) as x approaches infinity, if possible.

f(x)=
5x^2 - 3x + 7

a) h(x)= f(x)/x
b) h(x)= f(x)/x^2
c) h(x)= f(x)/x^3

Answer 1

a) make factors of given polynomial then divide by x, set limit and get answer

Answer 2

You factor out the largest term. So it would be x\(^2\)

Answer 3

where are you stuck?

Answer 4

so is it like this? f(x)= x^2(5) (3) +7 ?

Answer 5

\(\sf x^2[5-\frac{3}{x} +\frac{7}{x^2}]\)

Answer 6

and do I have to put all that over x?

Answer 7

for b) h(x) = 5 - 3/x + 7/x^2

Answer 8

so what do you think the limit of 5 - 3/x + 7/x^2 is, as x approaches infinity?

Answer 9

How do I solve that? am I supposed to take the square root of everything to cancel out?

Answer 10

no plug in x = infinity into h(x)

Answer 11

what does the value of 3/x approach as x approaches infinity?

Answer 12

it approaches 3+ right?

Answer 13

no:- 3 / infinity is close to what value?

Answer 14

can you see that its zero?

Answer 15

the same goes for 7/x^2
so limit as x --> infinity of 5 - 3/x + 7/x^2 is 5

Answer 16

I don't understand this!! :(

Answer 17

so was I setting a) up correctly?

Answer 18

in the case of a) h(x) = f(x)/x
so h(x) = 5x - 3 + 7/x
now you consider the limit of this as x ---> infinity

Answer 19

ok so I don't have to factor

Answer 20

yeah I have to find the limit of 5x-3+7

Answer 21

and that is 5x-3+7=0
5x+4= 0
x=4/5 ?

Answer 22

Turn the fraction into each of its individual terms, then find the limit of each term individually.

So maybe you have (6x^2-x)/(2x^2) then make it: (6x^2/2x^2)+(-x/2x^2) and see that you have 3-(1/2x) and if you're taking the limit as that approaches infinity, the first term isn't dependent on x, so you have 3+ something else. The 1/(2x) as that approaches infinity approaches 0, so your final answer will be 3+0, or 3 for this example.

Answer 23

ok

Answer 24

thnx