Question

Any quick way of finding the limit without l'Hopital rule?

Answer 1

\[\lim_{x \rightarrow \infty} \frac{ (3-n)^4-(2-n)^4 }{ (1-n)^4-(1+n)^4 }\]

Answer 2

u can use binomial exp by taking n out of the bracket

Answer 3

or try to visualize the leading coefficient of the numerator and denominator

Answer 4

\[\lim_{n \rightarrow \infty} \frac{ (n-3)^4-(n-2)^4 }{ (n-1)^4-(n+1)^4 }\]
both numerator and denominator are polynomials of degree 3, since \(n^2\) will add up to zero

Answer 5

is it 1/2

Answer 6

it is \(\frac{1}{2}\)
cube term in the numerator is \(-12n^3+8n^3=-4n^3\)

Answer 7

cube term in the denominator is \(-4n^3-4n^3=-8n^3\) and the ratio is \(\frac{1}{2}\)

Answer 8

So, if I expand first terms using binomial theorem I'll get\[\frac{ n^4-12n^3+...-(n^4-8n^3+...) }{ n^4-4n^3+...-(n^4+4n^3+...) } \approx \frac{ -4n^3 }{ -8n^3 }\]
and the limit will be 0.5?
Well, I guess it will be 0.5 since you already wrote it all lol

Answer 9

there are no \(n^4\) terms in this

Answer 10

see how short this method is
isnt it beautiful

Answer 11

oh yes, i see what you wrote

Answer 12

Hey the other way out is to write the terms as a^2 - b^2 = (a+b)(a-b).... That way you won't require this much simplification also...

Answer 13

i tried that but the algebra was annoying
found it easier to visualize the cube terms top and bottom, but that is a matter of taste

Answer 14

a good one @saloniiigupta95

Answer 15

it is all about mind set
what method gets fit where truly hard to guess

Answer 16

It gets you to the answer 1/2 easily... just u need to divide numerator and denominator by n smartly... @satellite73

Answer 17

@saloniiigupta95 and how should I do it smartly?

Answer 18

By smarrtly, I mean first opening it using \[a^2 - b^2 = (a+b)(a-b)\]