Matt wrote:
Need a bit of a hand with -shudders- Calculus.
Basically, I need to work out:
lim(n -> ∞) of n+1- √(n^2 – n+1)
We actually haven't done finding limits yet, so this is actually a bit of extension... I still know what a limit is though, but if you use any special theora, you'll need to explain them...
You will first need to multiply and divide by the conjugate, that is:
lim (n+1- √(n^2 – n+1)) * [(n+1 + √(n^2 – n+1)) / (n+1 + √(n^2 – n+1))]
Note we are essentially multiplying the limit by 1, so we are not actually changing anything. If you distribute it out you get:
lim [(n+1)^2 - (n^2 - n + 1)] / [n+1 + √(n^2 – n+1)]
If you distribute out the top further, you can cancel the n^2 terms, and this gives you:
lim (2n + 1 + n - 1)/(n+1 + √(n^2 – n+1))
The numerator simplifies further and you get:
lim (3n)/(n+1 + √(n^2 – n+1))
Now we need to use a theorem known as L'Hopital's Rule. It states:
If lim f(x)/g(x) yields infinity/infinity or 0/0 upon substitution,
Then lim f(x)/g(x) = lim f'(x)/g'(x). Meaning you can take the derivative of the numerator and divide by the derivative of the denominator and the answer remains unchanged.
Note for us, lim n->infinity of (3n)/(n+1 + √(n^2 – n+1)) = infinity/infinity, thus:
lim (3n)/(n+1 + √(n^2 – n+1)) = lim (3n)'/(n+1 + √(n^2 – n+1))'
I rather dislike using this approach, because if you haven't done limits, I don't think you would have done derivatives yet either. But this is the only way I can think of to solve this limit without using even more complicated theorems (e.g. Taylor expansion). There might be some algebraic trick that I'm not seeing after we multiply by the conjugate. Maybe someone else can spot it.
Anyways, if you do take the derivative you'll get:
lim 3/(1 + (2n-1)^(-1/2)) = 3/[1 + 1/sqrt(2n-1)] = 3/(1+0) = 3
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