# [SOLVED]lim as n to infty

#### dwsmith

##### Well-known member
Suppose $\lim\limits_{n\to\infty}a_n = a$. Prove using the delta-epsilon definition that $\lim\limits_{n\to\infty}a_n^2 = a^2$.

Let $\epsilon > 0$ be given. Let $N\in\mathbb{Z}$. Then for $n > N$, $|a_n - a| < \epsilon$.

Then
$$|a_n^2 - a^2| = |(a_n - a)(a_n + a)| < \epsilon |a_n + a|$$

I am stuck here.

#### Fantini

##### "Read Euler, read Euler." - Laplace
MHB Math Helper
I think you can argue as follows: consider

$$|a_n +a| = |a_n - a + 2a| \leq |a_n -a| + 2|a| < \varepsilon + 2|a|.$$

The result is trivial if $a =0$, therefore it's safe to assume $a \neq 0$.

Since $\varepsilon$ is arbitrary, you could say that $|a_n +a| \leq 2 |a|$.

Thus, take $N_0 \in \mathbb{N}$ such that for all $n \geq N_0$ we have $|a_n -a| < \frac{\varepsilon}{2 |a|}$.

It follows that for all $n \geq N_0$ we have

$$|a_n^2 -a^2| = |(a_n -a)(a_n +a)| \leq |(a_n -a)| \cdot 2 |a| < \frac{\varepsilon}{2 |a|} \cdot 2 |a| = \varepsilon.$$

Again, not entirely sure. Hope it helps, at least.

#### chisigma

##### Well-known member
Suppose $\lim\limits_{n\to\infty}a_n = a$. Prove using the delta-epsilon definition that $\lim\limits_{n\to\infty}a_n^2 = a^2$.

Let $\epsilon > 0$ be given. Let $N\in\mathbb{Z}$. Then for $n > N$, $|a_n - a| < \epsilon$.

Then
$$|a_n^2 - a^2| = |(a_n - a)(a_n + a)| < \epsilon |a_n + a|$$

I am stuck here.
Let's suppose to have two sequences $a_{n}$ and $b_{n}$ and that $\displaystyle \lim_{n \rightarrow \infty} a_{n}=a$ and $\displaystyle \lim_{n \rightarrow \infty} b_{n}=b$. That means that...

$\lim_{n \rightarrow \infty} (a_{n}-a)=\lim_{n \rightarrow \infty} (b_{n}-b)=0$ (1)

... and therefore, given a $\varepsilon>0$, it exists an $N$ for which $\forall n>N$ is...

$\displaystyle |(a_{n}-a)-0|< \sqrt{\varepsilon}$

$\displaystyle |(b_{n}-b)-0|< \sqrt{\varepsilon}$ (2)

... and then multiplying the (2) toghether...

$|(a_{n}-a)\ (b_{n}-b)-0|=|a_{n}-a|\ |b_{n}-b| < \varepsilon \implies \lim_{n \rightarrow \infty} (a_{n}-a)\ (b_{n}-b)=0$ (3)

Now we use (3) and the identity...

$\displaystyle a_{n}\ b_{n}= (a_{n}-a)\ (b_{n}-b) + b\ a_{n} + a\ b_{n} - a\ b$ (4)

... to obtain...

$\displaystyle \lim_{n \rightarrow \infty} a_{n}\ b_{n} = \lim_{n \rightarrow \infty} (a_{n}-a)\ (b_{n}-b) + \lim_{n \rightarrow \infty} b\ a_{n} + \lim_{n \rightarrow \infty} a\ b_{n} - \lim_{n \rightarrow \infty} a\ b = b\ a + a\ b - a\ b = a\ b$ (5)

In other word the limit of the product is the product of limits. In the particular case $b_{n}=a_{n}$ You have $\displaystyle \lim_{n \rightarrow \infty} a^{2}_{n}= a^{2}$...

Kind regards

$\chi$ $\sigma$

#### dwsmith

##### Well-known member
Let's suppose to have two sequences $a_{n}$ and $b_{n}$ and that $\displaystyle \lim_{n \rightarrow \infty} a_{n}=a$ and $\displaystyle \lim_{n \rightarrow \infty} b_{n}=b$. That means that...

$\lim_{n \rightarrow \infty} (a_{n}-a)=\lim_{n \rightarrow \infty} (b_{n}-b)=0$ (1)

... and therefore, given a $\varepsilon>0$, it exists an $N$ for which $\forall n>N$ is...

$\displaystyle |(a_{n}-a)-0|< \sqrt{\varepsilon}$

$\displaystyle |(b_{n}-b)-0|< \sqrt{\varepsilon}$ (2)

... and then multiplying the (2) toghether...

$|(a_{n}-a)\ (b_{n}-b)-0|=|a_{n}-a|\ |b_{n}-b| < \varepsilon \implies \lim_{n \rightarrow \infty} (a_{n}-a)\ (b_{n}-b)=0$ (3)

Now we use (3) and the identity...

$\displaystyle a_{n}\ b_{n}= (a_{n}-a)\ (b_{n}-b) + b\ a_{n} + a\ b_{n} - a\ b$ (4)

... to obtain...

$\displaystyle \lim_{n \rightarrow \infty} a_{n}\ b_{n} = \lim_{n \rightarrow \infty} (a_{n}-a)\ (b_{n}-b) + \lim_{n \rightarrow \infty} b\ a_{n} + \lim_{n \rightarrow \infty} a\ b_{n} - \lim_{n \rightarrow \infty} a\ b = b\ a + a\ b - a\ b = a\ b$ (5)

In other word the limit of the product is the product of limits. In the particular case $b_{n}=a_{n}$ You have $\displaystyle \lim_{n \rightarrow \infty} a^{2}_{n}= a^{2}$...

Kind regards

$\chi$ $\sigma$
Line (4) should be $+ab$. We would need a $-ab$ though which we don't have since we have a negative times a negative.

#### chisigma

##### Well-known member
Line (4) should be $+ab$. We would need a $-ab$ though which we don't have since we have a negative times a negative.
Why don't develop the expression...

$\displaystyle (a_{n}-a)\ (b_{n}-b) + a\ b_{n} + b\ a_{n} -a\ b$ (1)

... and observe what is the result?...

Kind regards

$\chi$ $\sigma$