Find the limit of the function using cancellation techniques (-6 + x) / x^4
Please explain.

- anonymous

Find the limit of the function using cancellation techniques (-6 + x) / x^4
Please explain.

- schrodinger

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- anonymous

I know how to do cancellation, but only with numerators that are factor-able.

- anonymous

Like Use cancellation techniques to evaluate limit as x approaches -1 of 2x squared minus x minus 3 divided by (x + 1).
You can factor the numerator as (2x -3)(x + 1). After cancelling out the common factor of (x+1), the limit can be evaluated using direct substation. The limit is equal to -5

- Luigi0210

@hartnn can help ya (:

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## More answers

- hartnn

separate the numerator
\(-6/x^4 \) + \(x/x^4\)
cancel out one x in the 2nd term

- anonymous

So \[-6/x^4 + x^3\]

- hartnn

-6/x^4 + 1/x^3

- anonymous

Why is 1 in the numerator?

- hartnn

|dw:1436550810803:dw|

- anonymous

Oh, that went right over my head.

- anonymous

So wouldn't it be 0 as the limit?

- hartnn

where does the x approach?
x-> 0 ?

- anonymous

I don't know. I was taught to just to insert whatever number, in this case 0, in the factored equation.

- hartnn

"in this case 0"
you deduced that from the question, right?

- anonymous

Yeah. Wait, wouldn't it not exist?

- hartnn

" just to insert whatever number"
is the worst way anyone can teach limits :P

- anonymous

It's virtual school. .-.

- hartnn

if x approaches 0,
the numerator is negative,
the denominator is very very near to 0
hence the limit will approach -infinity :)

- anonymous

Thats not an option...

- hartnn

***
negative number /0 = -infinity

- hartnn

what are the options?
can yo post the screenshot of entire question and options?

- anonymous

##### 1 Attachment

- hartnn

so they haven't introduced you to infinity yet...
in that case, you can choose "Does Not Exist" as the answer.
even though the limit exist and = - infinity

- anonymous

Exactly! It's basic math, but some websites say it would be 0, if and only if the x--> infinity.

- hartnn

when x-> infinity,
means x is a very large number
then 1/x will be very small number
hence, 1/x^3 and 1/x^4 will approach 0
hence you limit will be = 0

- anonymous

Yep. Thank you for the help! I really appreciated it. cx

- hartnn

welcome ^_^

- anonymous

It was that they did not exist. cx @hartnn

- hartnn

so we were correct :)

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