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I know how to do cancellation, but only with numerators that are factor-able.
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
@hartnn can help ya (:
separate the numerator \(-6/x^4 \) + \(x/x^4\) cancel out one x in the 2nd term
So \[-6/x^4 + x^3\]
-6/x^4 + 1/x^3
Why is 1 in the numerator?
Oh, that went right over my head.
So wouldn't it be 0 as the limit?
where does the x approach? x-> 0 ?
I don't know. I was taught to just to insert whatever number, in this case 0, in the factored equation.
"in this case 0" you deduced that from the question, right?
Yeah. Wait, wouldn't it not exist?
" just to insert whatever number" is the worst way anyone can teach limits :P
It's virtual school. .-.
if x approaches 0, the numerator is negative, the denominator is very very near to 0 hence the limit will approach -infinity :)
Thats not an option...
*** negative number /0 = -infinity
what are the options? can yo post the screenshot of entire question and options?
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
Exactly! It's basic math, but some websites say it would be 0, if and only if the x--> infinity.
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
Yep. Thank you for the help! I really appreciated it. cx
It was that they did not exist. cx @hartnn
so we were correct :)