## Limits

Last updates: 27 May 2012

## Limits

$l=\underset{x\to a}{\mathrm{lim}}f\left(x\right)$    means:    If $\epsilon \in {ℝ}_{>0}$ then there exists $\delta \in {ℝ}_{>0}$ such that if $d\left(x,a\right)<0$ then $d\left(f\left(x\right),l\right)<\epsilon$.

$l=\underset{n\to \infty }{\mathrm{lim}}{a}_{n}$    means:    If $\epsilon \in {ℝ}_{>0}$ then there exists $N\in {ℤ}_{>0}$ such that if $n\in {ℤ}_{>0}$ and $n>N$ then $d\left({a}_{n},l\right)<\epsilon$.

Let $X$ be a metric space. Let $f:X\to ℝ$ and $g:X\to ℝ$ be functions and let $a\in X$. Assume that $\underset{x\to a}{\mathrm{lim}}f\left(x\right)$ and $\underset{x\to a}{\mathrm{lim}}g\left(x\right)$ exist. Then

1. (a) $\underset{x\to a}{\mathrm{lim}}\left(f\left(x\right)+g\left(x\right)\right)=\underset{x\to a}{\mathrm{lim}}f\left(x\right)+\underset{x\to a}{\mathrm{lim}}g\left(x\right)$,
2. (b) If $c\in ℝ$ then $\underset{x\to a}{\mathrm{lim}}cf\left(x\right)=c\underset{x\to a}{\mathrm{lim}}f\left(x\right)$,   and
3. (c) $\underset{x\to a}{\mathrm{lim}}f\left(x\right)g\left(x\right)=\left(\underset{x\to a}{\mathrm{lim}}f\left(x\right)\right)\left(\underset{x\to a}{\mathrm{lim}}g\left(x\right)\right)$.

 Proof of part (a).

 Proof of part (b):

 Proof of part (c):

Assume that $\underset{x\to a}{\mathrm{lim}}g\left(x\right)=l$ and $\underset{y\to l}{\mathrm{lim}}f\left(y\right)$ exists. Then $limy→l f(y)= limx→a f(g(x)).$

 Proof.

Let ${a}_{n}$ and ${b}_{n}$ be sequences in $ℝ$. Assume that $\underset{n\to \infty }{\mathrm{lim}}{a}_{n}$ exists and $\underset{n\to \infty }{\mathrm{lim}}{b}_{n}$ exists. $If an≤bn then limn→∞ an ≤ limn→∞ bn.$

 Proof.

## The functions $f\left(x\right)={x}^{n}$ and $f\left(x\right)={e}^{x}$

1. (a) Let $n\in {ℤ}_{>0}$ and $a\in ℝ$. Then ${x}^{n}$ is continuous at $x=a$ (i.e. $\underset{x\to a}{\mathrm{lim}}{x}^{n}={a}^{n}$).
2. (b) ${e}^{x}$ is continuous at $x=a$.
3. (c) Prove that $\underset{x\to 0}{\mathrm{lim}}{e}^{x}={e}^{0}.$
4. (d) Prove that $\underset{x\to a}{\mathrm{lim}}{e}^{x}={e}^{a}$.

 Proof of part (a).

 First part of proof of part (b).

 Proof of part (d).

1. (a) Let $x\in ℂ$. $limn→∞ xn ={ 0, if |x|<1, diverges, if |x|>1, 1, ifx=1, diverges, if |x|=1 and x≠1,$
2. (b) Let $x\in ℂ$. $limn→∞ 1+x+x2+ ⋯+xn = limn→∞ 1-xn+ 11-x ={ 11-x, if |x|<1, diverges, if|x| ≥1.$

 Proof of part (a) Case $|x|<1$.

 Proof of part (a) Case $|x|>1$.

 Proof of part (b).

## Useful limits

1. (a) If $n\in {ℤ}_{>0}$ then $\underset{x\to \infty }{\mathrm{lim}}{x}^{n}{e}^{-x}=0$.
2. (b) If $\alpha \in {ℝ}_{>0}$ then $\underset{x\to \infty }{\mathrm{lim}}{x}^{-\alpha }\mathrm{log}x=0$.
3. (c) Let $p\in {ℝ}_{>0}$. Then $\underset{n\to \infty }{\mathrm{lim}}\frac{1}{{n}^{p}}=0$.
4. (d) Let $p\in {ℝ}_{>0}$. Then $\underset{n\to \infty }{\mathrm{lim}}{p}^{1/n}=1$.
5. (e) $\underset{n\to \infty }{\mathrm{lim}}{n}^{1/n}=1$.
6. (f) Let $\alpha \in ℝ$ and $p\in {ℝ}_{>0}$. Then $\underset{n\to \infty }{\mathrm{lim}}\frac{{n}^{\alpha }}{{\left(1+p\right)}^{n}}=0$.
7. (g) If $|x|<1$ then $\underset{n\to \infty }{\mathrm{lim}}{x}^{n}=0$.
8. (h) If $0<\alpha <1$ then $\underset{x\to 0}{\mathrm{lim}}{x}^{\alpha }=0$.
9. (i) $\underset{x\to 0}{\mathrm{lim}}\frac{{e}^{x}-1}{x}=1$.
10. (j) $\underset{x\to 0}{\mathrm{lim}}\frac{\mathrm{sin}x}{x}=1$.
11. (k) $\underset{x\to 0}{\mathrm{lim}}\frac{\mathrm{cos}x-1}{{x}^{2}}=-\frac{1}{2}$.
12. (l) $\underset{x\to 0}{\mathrm{lim}}\frac{\mathrm{log}\left(1+x\right)}{x}=1$.

 Proof of part (l): $limx→0 log(1+x) x = limy→0 log(1+ ey-1) ey-1 = limy→0 yey-1 = limy→0 1 ( ey-1 y ) = 11 = 1.$ $\square$

 Proof of part (j): $limx→0 sinxx = limx→0 eix -e-ix 2ix = limx→0 eix-1 2ix - e-ix-1 2ix = limx→0 12 ( eix-1 ix ) + 12 ( e-ix-1 -ix ) = 12·1 +12·1 = 1.$ $\square$

 Proof of part (a): Assume $n\in {ℤ}_{>0}$. $0≤ limx→∞ xne-x = limx→∞ xnex ≤ limx→∞ xn 1 (n+1)! xn+1 = limx→∞ (n+1)!x = (n+1)! limx→∞ 1x = 0.$ $\square$

 Proof of part (b): Let $0<\epsilon <\alpha$. $0≤ limx→∞ x-α logx = limx→∞ (x-α ∫1x 1tdt) ≤ limx→∞ (x-α ∫1xt ε-1dt) = limx→∞ x-α ( xε-1ε ε ) = limx→∞ ( xε-α -x-α ε) ≤ limx→∞ xε-α ε = 0.$ $\square$

 Proof of part (c). To show: If $\epsilon \in {ℝ}_{>0}$ then there exists $N\in {ℤ}_{>0}$ such that if $n\in {ℤ}_{>0}$ and $n>N$ then $|\frac{1}{{n}^{p}}|<\epsilon$. Assume $\epsilon \in {ℝ}_{{>}_{0}}$. Let $N={\left(\frac{1}{\epsilon }\right)}^{\frac{1}{p}}$. To show: If $n\in {ℤ}_{>0}$ and $n>N$ then $|\frac{1}{{n}^{p}}|<\epsilon$. Assume $n\in {ℤ}_{>0}$ and $n>N$. To show: $|\frac{1}{{n}^{p}}|<\epsilon$. $|1np| < |1Np| = 1( (1ε) 1p) p = 1(1ε) = ε.$ $\square$

 Proof of part (d):

 Proof of part (e):

## Notes and References

This section proves the fundamental limit theorems used most often for computations. THe presentation is focused on Theorem 1.1, establishing, without much effort, the primary tools that are used for evaluation of limits in examples.

It feels a bit funny not to include the f(x)/g(x) case in Theorem 1.1. SHOULD IT BE PUT IN? WHY? WHY NOT?

It is interesting to note that in practice, once Theorems 1.1 and 1.2 are proved, it is hardly ever necessary to use an $\epsilon \text{-}\delta$ proof to evaluate a limit.

A traditional reference for this material is [Ru1, Ch. 3 and 4] with Theorem 1.1 being [Ru1, Theorem 4.4] and much of Theorem ??? coinciding with [Ru1, Theorem 3.20].

## References

[Bou] N. Bourbaki, Algèbre, Chapitre ?: ??????????? MR?????.

[Ru2] W. Rudin, Real and complex analysis, Third edition, McGraw-Hill, 1987. MR0924157.