Question

Find the limit of the function algebraically.

limit as x approaches zero of quantity x squared plus three divided by x to the fourth power.

Answer 1

@genius12 @amistre64 please help me

Answer 2

\[\lim_{x \rightarrow 0}(x^2+3)/x^4 \]

Answer 3

theres nothing to cancel out top to bottom ... so this goes into infinity

Answer 4

couldn't you factor and cancel?

Answer 5

factor how?

Answer 6

maybe split ...

Answer 7

\[\frac{x^2}{x^4}+\frac{3}{x^4}\]
but that still leaves zeros underneath

Answer 8

Let me get an example from the lesson of how they did it please. They had one that would have substituted to 0/0 but they canceled and got it to 1/4

Answer 9

\[\lim_{x \rightarrow 2}(x-2)/(x^2-4)\]

Answer 10

they factored the bottom, to (x+2)(x-2) then cancled the x-2 and it was 1/x+2 =1/4

Answer 11

@amistre64 ?

Answer 12

your given equation does not cancel anything ... we remian left with an x in the denominator no matter what we do

Answer 13

an x in the bottom means that we run off into infinity ... the only question is is it the same infinity on each side of 0 ?

Answer 14

since x^4 or x^2 anything is positive or zero .... id say this runs up to +inf; but depending on the context that could just as well mean no limit exists

Answer 15

There is no way to factor the denominator, therefore it will always =0 so there is no limit.

That is what I submitted as my answer...is that correct or did it need more detail... I didn't think it could simplify any more just a bit confused since didn't have any that didn't simplify in my lesson practice.

Answer 16

im not the one grading it, but that sounds like youve got the idea of it. If you pursue mathematics they will expect more rigourous and "to the definition" answers tho.

Answer 17

I understand, and I am so can you please give me a more indepth answer then? I understand the basics of this topic but not well enough to go into great detail without concern of straying from the correct answer...I appreciate your help

Answer 18

if the limit as x approach a from the left is equal to the limit as x approached a from the right is the same; then the limit exists. Depending on the course, they will define limits to infinity in different ways.

let x = -a
\[\lim_{a\to 0}~\frac{1}{(-a)^2}+\frac{3}{(-a)^4}\]
\[\lim_{a\to 0}~\frac{1}{a^2}+\frac{3}{a^4}=\infty\]

let x = a
\[\lim_{a\to 0}~\frac{1}{(a)^2}+\frac{3}{(a)^4}\]
\[\lim_{a\to 0}~\frac{1}{a^2}+\frac{3}{a^4}=\infty\]

since the limit from the left and right are the same .... then some courses would accept infinity

but since the limit from the left and right are not finite .... then some courses would say it does not exist

Answer 19

OK. That makes a lot more sense and clears everything up for me. Thank you so much for your time.

Answer 20

good luck

Answer 21

Thanks I am going to need it this coming year!