Question

help me solve this.
lim 2x^3 +7x/3sin^2x
x - 0

Answer 1

@ganeshie8

Answer 2

@iambatman

Answer 3

@mathmale

Answer 4

\[\lim_{x\to0}\frac{2x^3+7x}{3\sin^2(x)}\]??

Answer 5

yes

Answer 6

then what will i do

Answer 7

probably has no limit

Answer 8

how it happen

Answer 9

the lim of thatequation approaches 0

Answer 10

what tools can you use? can you use l'hopital's rule?

Answer 11

and you are sure it is \(x^3\) up top right?

Answer 12

YES IT IS

Answer 13

ok
do you know l'hopital's rule?

Answer 14

no.

Answer 15

oh
what are you supposed to use to solve this then?

Answer 16

substitute 0 to the given equation?

Answer 17

lol no
that only tells you that you get \(\frac{0}{0}\) which is not a number
just means you need to do more work

Answer 18

how. what will i use

Answer 19

i take it this is a calculus course right?

Answer 20

yes it is.

Answer 21

but you did not get to L'Hopital's rule yet
we can try some algebra

Answer 22

\[\frac{2x^3+7x}{3\sin^2(x)}=\frac{x}{3\sin(x)}\times \frac{2x^2+7}{\sin(x)}\]

Answer 23

yes i didnt

Answer 24

\[\lim_{x\to 0}\frac{x}{3\sin(x)}=\frac{1}{3}\] since
\[\lim_{x\to 0}\frac{\sin(x)}{x}=1\]

Answer 25

how the 2x^2+7 and 3sin be cancel?

Answer 26

so we have
\[\frac{1}{3}\lim_{x\to 0}\frac{2x^2+7}{\sin(x)}\] to consider

Answer 27

now if you replace \(x\) by \(0\) in
\[\frac{2x^2+7}{\sin(x)}\] you do not get \(\frac{0}{0}\) but rather you get \(\frac{7}{0}\)
this means it is not "undetermined" i.e. you have more work to do
this means the limit does not exist

Answer 28

so theres no answer but it couldnt happen?

Answer 29

the same way
\[\lim_{x\to 0}\frac{1}{x}\] is undefined

Answer 30

yes. but it cant happen there must be an answer

Answer 31

@satellite73

Answer 32

a limit does not necessarily exist, so no, there need not be an answer