Limit question from calculus. Problem in comments
divide top and bottom by n^3
After I do that what do I do next?
well pretend n is a really large number... what does the expression equal?
0? Is it nonexistant?
no... are you in precalculus or calculus?
it's a good idea to get a feeling for intuition here
calculus lol. I'm taking an online course and I have no idea what I'm doing
you're looking at \(\Huge \frac{3-5/n^2}{1-2/n+1/n^3}\) as n becomes really large
that's what you get after you divide both sides by n^3
Yeah I get that but I have to pick one of those answers
if you're going to always worry about which answer to pick you'll have a lot of answers to pick for your course... honestly if you wanted the answer you could've just looked it up on wolfram alpha
anyways you're interested in what happens when n becomes really, really big
But I want to understand how to do it while getting the right answer.
so when n becomes really really big, some of the terms in that rational expression don't matter...
can you identify which ones?
by don't matter I mean that they approach zero.
The denominator numbers?
well, let's try it out. Let's take n=1 first
\(\Huge \frac{3-5/n^2}{1-2/n+1/n^3}\)
-2/0 is what I got
\(\Huge \frac{3-5/1^2}{1-2/1+1/1^3}\)=(3-5)/(1-2+1)=-2/((0)=undef
let's try it for n=10
\(\Huge \frac{3-5/10^2}{1-2/10+1/10^3}\)=3.6829
try it for n=100: we get 3.060711
try it for n=1000: we get 3.00600701102
try for n=10000: we get 3.00060007
The higher to infinity we go it gets closer to 3. So as n approaches infinity we get 3
yes... now, let's see why
as n becomes larger and larger, the terms with n^2 and n in the denominator become closer and closer to zero
as n approaches infinity, these terms approach zero
and we are simply left with 3/1=3
That makes so much sense now! Thank you so much.
np... just remember the rules you were taught in class: you have to divide by the term of the largest degree to be able to approach the problem like I did
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