Definition 2.1.1 (Limit of a sequence). Given a sequence of real numbers \(\{a_n\}\), we say \(\lim \limits _{n\to \infty } a_n = L\) if the following holds: \begin {align*} \forall \epsilon >0, \exists N>0 \text { such that if } n> N, \text {then } |a_n -L| < \epsilon \end {align*}
We say the Limit Exists for the sequence \(\{a_n\}\), or \(\{a_n\}\) converges, if there exists \(L \in \mathbb {R}\) such that \(a_n \to L\)
We start with the arithmetic properties of Limits and prove them with the formal definition of limits given above(2.1.1)
Statement:
If \(\{a_n\}\) is a sequence of positive numbers such that \(\lim \limits _{n \to \infty }a_n =L\), we need to prove using \(\epsilon , N\) definition that \begin {align*} \lim \limits _{n \to \infty }a_n^p = L^p \end {align*}
Where \(p \in \mathbb {N}\)
We need to show that for any \(\epsilon >0\) there exists some \(N \in \mathbb {N}\) such that if \(n>N\) \begin {align*} |a_n^p-L^p| < \epsilon \end {align*}
observations:
Using the identity
\begin {align*} x^p-y^p = (x-y)(x^{p-1}+x^{p-2}y+...+xy^{p-2}+y^{p-1}) \end {align*}
we have \begin {align*} |a_n^p -L^p| = |a_n-L||a_n^{p-1}+a_n^{p-2}L+...+a_nL^{p-2}+L^{p-1}|\\ \implies |a_n^p -L^p| \leq |a_n-L|||a_n^{p-1}|+|a_n^{p-2}L|+...+|a_nL^{p-2}|+|L^{p-1}|| \end {align*}
We need to bound \(|a_n^kL^m|\) where \(k,m \in \mathbb {N}\) and \(0\leq k,m \leq p-1\) with \(k+m =p-1\). \begin {align*} |a_n-L| <1 &\implies -1 < a_n-L<1 \implies L-1<a_n<L+1\\ &\implies |a_n| < |L|+1 \qquad \text { for some } n > N_1\\ &\implies |a_n|^k |L|^m <(|L|+1)^k|L|^m \end {align*}
for elegance, let \begin {align*} M = \max \{|L|^{p-1},(|L|+1)^1|L|^{p-2},...,(|L|+1)^{p-2}|L|^{1},(|L|+1)^{p-1}\}\\ \implies |a_n^p -L^p| < |a_n-L|pM \end {align*}
Hence we bound \(|a_n^p-L^p|\) above by \(\epsilon \) by requiring that: \begin {align*} |a_n-L| < \frac {\epsilon }{pM} \end {align*}
for some \(p \in \mathbb {N}\).
Proof.
Now, we can complete the proof. Because \(a_n \to L\), \(\exists N_1 \in \mathbb {N}\) such that if \(n > N_0\) then for \(\epsilon _0 = 1\), \begin {align*} |a_n-L| <1 &\implies |a_n| < |L|+1 \\ &\implies |a_n|^k |L|^m <(|L|+1)^k|L|^m \end {align*}
For \(k,m \in \mathbb {N}\) and \(0\leq k,m \leq p-1\) with \(k+m =p-1\).
Again, using the fact that \(a_n \to L\), \(\exists N_1 \in \mathbb {N}\) such that if \(n > N_1\) then for \(p \in \mathbb {N}\) and some \(\epsilon _1\), \begin {align*} |a_n-L| < \epsilon _1 = \frac {\epsilon }{pM} \end {align*}
Then, for any arbitrary \(\epsilon > 0\), there exists \(N = max\{N_0, N_1\}\) such that if \(n > N\), both the previous inequalities hold: \begin {align*} |a_n^p -L^p| = |a_n-L||a_n^{p-1}+a_n^{p-2}L+...+a_nL^{p-2}+L^{p-1}| \\ \implies |a_n^p -L^p| \leq |a_n-L|\left ||a_n^{p-1}|+|a_n^{p-2}L|+...+|a_nL^{p-2}|+|L^{p-1}|\right |\\ \implies |a_n^p -L^p| < |a_n-L|pM\\ \implies |a_n^p -L^p| < \frac {\epsilon }{pM} * pM = \epsilon \end {align*}
Hence, if \(a_n\to L\), \(a_n^p \to L^p\) for positive sequences of \(a_n\) and \(p \in \mathbb {N}\) □
Statement:
If \(\{a_n\}\) is a sequence of positive numbers such that \(\lim \limits _{n \to \infty } a_n = L\), then \begin {align*} \lim \limits _{n\to \infty } a_n^{\frac {p}{q}} = L^{\frac {p}{q}} \end {align*}
Or equivalently, for any \(\epsilon >0\), \(\exists N \in \mathbb {N}\) such that for all \(n>N\) we have: \begin {align*} |a_n^{\frac {p}{q}} - L^{\frac {p}{q}}| &< \epsilon \\ |a_n^{\frac {p}{q}} - L^{\frac {p}{q}}| &= |(a_n^p)^{\frac {1}{q}}-(L^p)^{\frac {1}{q}}|\\ \implies |a_n^{\frac {p}{q}} - L^{\frac {p}{q}}| &= \frac {|a_n^p -L^p|}{|(L^p)^{\frac {q-1}{q}}+ (a_n^p)^{\frac {1}{q}}(L^p)^{\frac {q-2}{q}}+...+(a_n^p)^{\frac {q-2}{q}}(L^p)^{\frac {1}{q}}+(a_n^p)^{\frac {q-1}{q}}|} \leq \frac {|a_n^p -L^p|}{|(L^p)^{\frac {q-1}{q}}|} \end {align*}
Hence we must show that: \begin {align*} \frac {|a_n^p -L^p|}{|(L^p)^{\frac {q-1}{q}}|} < \epsilon \\ \implies |a_n^p -L^p| < |(L^p)^{\frac {q-1}{q}}|\epsilon \end {align*}
From 4 (a) we know that for positive sequence \(\{a_n\}\) that has a limit \(L\), \(a_n^p \to L^p\). Hence we can start the proof.
Proof. Given that \(a_n \to L\) and \(a_n\) is positive, \begin {gather*} a_n^p\to L^p \\ \therefore \forall \epsilon _1 > 0 \quad \exists N_1 \in \mathbb {N} \text { s.t for } n > N_1 \\ |a_n^p - L^p| < \epsilon _1 \end {gather*} Hence, choosing \(\epsilon _1 = |(L^p)^{\frac {q-1}{q}}|\epsilon \), where \(\epsilon \) is some arbitrary positive value, we can complete the proof. By taking \(N = N_1\) so that when \(n> N\), \begin {align*} |a_n^{\frac {p}{q}} - L^{\frac {p}{q}}| \leq \frac {|a_n^p -L^p|}{|(L^p)^{\frac {q-1}{q}}|} = \frac {\epsilon }{|(L^p)^{\frac {q-1}{q}}|} \times |(L^p)^{\frac {q-1}{q}}| = \epsilon \end {align*}
Hence, \(a_n^{\frac {p}{q}} \to L^{\frac {p}{q}}\) □
Statement:
If \(\{a_n\}\) is a sequence of real numbers such that \(\lim \limits _{n \to \infty }a_n =L \), \(\lim \limits _{n \to \infty }|a_n| =|L| \).
We need to prove that for any \(\epsilon >0\), \(\exists N \in \mathbb {N}\) such that for \(n > N\) \begin {align*} ||a_n|-|L|| < \epsilon \end {align*}
Note that, by the reverse triangle inequality (??) \begin {gather*} ||a_n|-|L|| \leq |a_n-L| \end {gather*} But, because \(a_n \to L\), it is indeed the case for any arbitrary \(\epsilon >0\), \(\exists N \in \mathbb {N}\) such that for all \(n >N\) \begin {align*} |a_n -L| < \epsilon \\ ||a_n|-|L|| \leq |a_n-L| < \epsilon \\ \implies ||a_n|-|L|| < \epsilon \end {align*}
Where \(a_k\) and \(b_k\) are the leading coefficients of \(p(x)\) and \(q(x)\) respectively.
\begin {align*} \lim _{n \to \infty } \dfrac {p(n)}{q(n)} & = \lim _{n \to \infty } \dfrac {a_kn^k+a_{k-1}n^{k-1}+ ... + a_1n+a_0}{b_kn^k+b_{k-1}n^{k-1}+ ... + b_1n+b_0}\\ & = \lim _{n \to \infty } \dfrac { \dfrac {a_kn^k+a_{k-1}n^{k-1}+ ... + a_1n+a_0}{n^k}}{\dfrac {b_kn^k+b_{k-1}n^{k-1}+ ... + b_1n+b_0}{n^k}} \\ &= \lim _{n \to \infty } \dfrac {a_k+\frac {a_{k-1}}{n}+ ... + \frac {a_1}{n^{k-1}}+\frac {a_0}{n^k}}{b_k+\frac {b_{k-1}}{n}+ ... + \frac {b_1}{n^{k-1}}+\frac {b_0}{n^k}} \end {align*}
since \(\lim \limits _{n \to \infty }\dfrac {1}{n^p} =0\) for \(p >0\)