# Use formal definition of a limit of a sequence to prove the following:

a) lim $$(n+1)/(n+2)=1$$ as n approaches infinity

b) lim $$(n^{3}-1)/(n^{4}-1)=0$$ as n approaches infinity

c) lim $$n!/n^{n}=0$$ as n approaches inifinity

I tried proving those but I'm not sure if i did it right so would appreciate if anyone could point out if i missed any steps or if I can improve them. Thank you

a) |[(n+1)/(n+2)]-1| = |[(n+1)-(n+2)]/(n+2)| = 3/(n+2)

n+2>n

1/(n+2)<1/n

3/(n+2)<3/n

So for any epsilon>0, 3/epsilon>0

By Archimedean property, there exists N in natural numbers such that N>3/epsilon

When n≥N, n≥N>3/epsilon

which implies that 1/n≤1/N<1/(3/epsilon) implying 3/n≤3/N<3/(3/epsilon)

So, $$|x_{n}-L|=3/(n+2)$$<3/n≤3/N<3/(3/epsilon)=epsilon

Therefore, $$|x_{n}-L|$$=3/(n+2)

implying that $$|x_{n}-L|$$

Then by definition of limit, $$x_{n}$$=(n+1)/(n+2) converges to real number and its limit is 0.

b)for any n≤N in natural numbers, $$n^{3}-1 for any n>2, we have 0<$$n^{4}-n^{3}$$<$$n^{4}-1$$

so 1/($$n^{4}-1$$)<1/($$n^{4}-n^{3}$$). Hence, for n>2, 0<($$n^{3}-1$$)/($$n^{4}-1$$)<$$n^{3}$$/($$n^{4}-1)$$<$$n^{3}$$/($$n^{4}-n^{3}$$)=1/(n-1)

So given any epsilon>0, it'll be sufficient to choose N such that 1/(N-1)((1/epsilon)+1)

Given epsilon>0, choose N=(1/epsilon)+1. Then N>(1/epsilon)+1, hence 1/(N-1)2, given any n≥N, we have 0<$$(n^{3}-1)/(n^{4}-1)$$<$$n^{3}/(n^{4}-1)$$<$$n^{3}/(n^{4}-n^{3})$$=1/(n-1)<1/(N-1)

Therefore, the stated limit is indeed 0.

c) |n!/$$n^{n}$$|<1/n for all n in Natural numbers excluding 0

for any epsilon>0, 1/epsilon>0, there exists N in Natural numbers such that N>1/epsilon, when n≥N, n≥N>1/epsilon

implying that 1/n≤1/N<1/(1/epsilon), therefore, 1/n≤1/N

So, |$$x_{n}-L$$|=n!/$$n^{n}$$<1/n≤1/N

Therefore, |$$x_{n}-L$$|

By definition of a limit, $$x_{n}=n!/n^{n}$$ converges to real number and its limit is 0.

• Please use mathjax to typeset any math content. – Viktor Glombik Jan 26 at 19:47