Proving limits by definition II

Limit of functions

Limit of functions is a concept that is used to describe the behavior of a function as it approaches a certain point.

General definition

$$\mathrm{\forall }\epsilon >0\mathrm{\exists }\delta >0\mathrm{\forall }x\in D_f,0<|x-a|<\delta :|f(x)-L|<\epsilon$$

Which means that for every positive number $\epsilon$, there is a positive number $\delta$ such that for every $x$ in the domain of $f$, if $0<|x-a|<\delta$, then $|f(x)-L|<\epsilon$.

Specific cases

$A=\underset{a}{lim}f$	$A\in \mathbb{R}$	$A=+\mathrm{\infty }$	$A=-\mathrm{\infty }$
$a\in \mathbb{R}$	finite limit at a finite point	positively infinite limit at a finite point	negatively infinite limit at a finite point
$a=+\mathrm{\infty }$	finite limit at positive infinity	positively infinite limit at positive infinity	negatively infinite limit at positive infinity
$a=-\mathrm{\infty }$	finite limit at negative infinity	positively infinite limit at positive infinity oxygen	negatively infinite limit at negative infinity

For each of these cases, the definition of the limit is the same, but the way we write it is different.

Limits at finite points

Finite limit at a finite point

$$\underset{x\to a}{lim}f(x)=A$$$$\mathrm{\forall }\epsilon >0\mathrm{\exists }\delta >0\mathrm{\forall }x\in D_f,0<|x-a|<\delta :|f(x)-A|<\epsilon$$

Positively infinite limit at a finite point

$$\underset{x\to a}{lim}f(x)=+\mathrm{\infty }$$$$\mathrm{\forall }P>0\mathrm{\exists }\delta >0\mathrm{\forall }x\in D_f,0<|x-a|<\delta :f(x)>P$$

Negatively infinite limit at a finite point

$$\underset{x\to a}{lim}f(x)=-\mathrm{\infty }$$$$\mathrm{\forall }P<0\mathrm{\exists }\delta >0\mathrm{\forall }x\in D_f,0<|x-a|<\delta :f(x)<P$$

Limits at positive infinity

Finite limit at positive infinity

$$\underset{x\to +\mathrm{\infty }}{lim}f(x)=A$$$$\mathrm{\forall }\epsilon >0\mathrm{\exists }R>0\mathrm{\forall }x\in D_f,x>R:|f(x)-A|<\epsilon$$

Positively infinite limit at positive infinity

$$\underset{x\to +\mathrm{\infty }}{lim}f(x)=+\mathrm{\infty }$$$$\mathrm{\forall }P>0\mathrm{\exists }R>0\mathrm{\forall }x\in D_f,x>R:f(x)>P$$

Negatively infinite limit at positive infinity

$$\underset{x\to +\mathrm{\infty }}{lim}f(x)=-\mathrm{\infty }$$$$\mathrm{\forall }P<0\mathrm{\exists }R>0\mathrm{\forall }x\in D_f,x>R:f(x)<P$$

Limits at negative infinity

Finite limit at negative infinity

$$\underset{x\to -\mathrm{\infty }}{lim}f(x)=A$$$$\mathrm{\forall }\epsilon >0\mathrm{\exists }R<0\mathrm{\forall }x\in D_f,x<R:|f(x)-A|<\epsilon$$

Positively infinite limit at negative infinity

$$\underset{x\to -\mathrm{\infty }}{lim}f(x)=+\mathrm{\infty }$$$$\mathrm{\forall }P>0\mathrm{\exists }R<0\mathrm{\forall }x\in D_f,x<R:f(x)>P$$

Negatively infinite limit at negative infinity

$$\underset{x\to -\mathrm{\infty }}{lim}f(x)=-\mathrm{\infty }$$$$\mathrm{\forall }P<0\mathrm{\exists }R<0\mathrm{\forall }x\in D_f,x<R:f(x)<P$$

One-sided limits

One-sided limits are used to describe the behavior of a function as it approaches a certain finite point from one side.

Some functions may have different limits from the left and from the right, depending on whether the function is continuous at that point.

Example

Regular limit

$$\mathrm{\exists }\underset{x\to a}{lim}f=A\iff \mathrm{\exists }\underset{x\to a^{+-}}{lim}f=A\iff \underset{x\to a^{-}}{lim}f=\underset{x\to a^{+}}{lim}f=A$$

One-sided limit

$$\underset{x\to a^{-}}{lim}f=A\ne \underset{x\to a^{+}}{lim}f=B$$

Example 1 with extra commentary

Prove by the definition of the limit that:

$$\underset{x\to 0}{lim}\frac{1}{1+x}=1$$$$f(x)=\frac{1}{1+x}$$$$D_f=\mathbb{R}\setminus \{-1\}$$

By definition

$$\mathrm{\forall }\epsilon >0\mathrm{\exists }\delta >0\mathrm{\forall }x\in D_f=\mathbb{R}\setminus \{-1\},0<|x-0|<\delta :|\frac{1}{1+x}-1|<\epsilon$$

We need to prove that for every positive number $\epsilon$, there exists such a positive number $\delta$ that if the distance between $x$ and $0$ is less than $\delta$, then the distance between $f(x)$ and $1$ is less than $\epsilon$.

Proof

Fix $\epsilon >0$.

Find $\delta >0$ such that $|\frac{1}{1+x}-1|<\epsilon$ when $0<|x-0|<\delta$.

So we have this implication

$$0<|x-0|<\delta ⟹|\frac{1}{1+x}-1|<\epsilon$$

To figure out what $\delta$ we need, let's try to transform $\epsilon$ inequality so that $|x-0|$ is factored out, then we would see the connection between $\delta$ and $\epsilon$.

$$|\frac{1}{1+x}-1|=\left|\frac{1-1-x}{1+x}\right|=\left|\frac{-x}{1+x}\right|=|x|\cdot \frac{1}{|1+x|}<\epsilon$$

Now we can see that essentially, what we want to achieve is

$$|x|\cdot \frac{1}{|1+x|}<\delta \cdot \underset{g(x)}{\underbrace{\frac{1}{|1+x|}}}$$

But we cannot make $\delta$ dependant on $x$, since delta is always a function of $\epsilon$ only.

Therefore, if we can find some constant $C$ such that $\frac{1}{|1+x|}<C$ then $\delta \le \frac{\epsilon }{C}$ would work.

Since we choose $\delta$ and $|x-0|<\delta ⟹\frac{1}{|1+x|}$ cannot be arbitrarily large, so we can find such $C$.

So let's put an arbitrary small restriction on $\delta$ and say that $\delta \le \frac{1}{2}$. (We would usually choose $1$ for simplicity, but since $-1$ is not in the domain, let's take a smaller number).

Now if $\delta \le \frac{1}{2}$, then

$$0<|x-0|<\delta ⟹0<|x|<\frac{1}{2}$$

Or in another form

$$-\frac{1}{2}<x<\frac{1}{2}$$

Now let's make this inequality work for $g(x)=\frac{1}{|1+x|}$.

$$-\frac{1}{2}<x<\frac{1}{2}⟹\frac{1}{2}<1+x<\frac{3}{2}⟹$$$$⟹\frac{1}{2}<|1+x|<\frac{3}{2}⟹\frac{2}{3}<\frac{1}{|1+x|}<2$$

So within our constraints, we found that $g(x)$ is at most $2$.

Then

$$2\cdot |x|<\epsilon ⟹|x|<\frac{\epsilon }{2}$$

We need delta to satisfy two conditions:

• $\delta \le \frac{1}{2}$
• $\delta \le \frac{\epsilon }{2}$

Easiest way to do it is to just choose $\delta =min\{\frac{1}{2},\frac{\epsilon }{2}\}$.

And now we can finally state that

$$\mathrm{\forall }\epsilon >0\mathrm{\exists }\delta =min\{\frac{1}{2},\frac{\epsilon }{2}\}\mathrm{\forall }x\in D_f=\mathbb{R}\setminus \{-1\},0<|x-0|<\delta :|\frac{1}{1+x}-1|<2\cdot |x-0|\le \epsilon$$

Or in short

$$\delta =min\{\frac{1}{2},\frac{\epsilon }{2}\}⟹|\frac{1}{1+x}-1|<\epsilon \text{when}0<|x-0|<\delta$$

Example 2

Prove by the definition of the limit that:

$$\underset{x\to 1}{lim}\frac{x^4+2x^2-3}{x^2-3x+2}=-8$$$$f(x)=\frac{x^4+2x^2-3}{x^2-3x+2}=\frac{x^4+2x^2-3}{(x-1)(x-2)}$$$$D_f=\mathbb{R}\setminus \{1,2\}$$

Note that $x\to 1$, numerator might also have a factor of $(x-1)$, try to divide by it using Horner's method.

$$f(x)=\frac{x^4+2x^2-3}{(x-1)(x-2)}=\frac{\cancel{(x-1)}(x^3+x^2+3x+3)}{\cancel{(x-1)}(x-2)}=\frac{x^3+x^2+3x+3}{x-2}$$

By definition

$$\mathrm{\forall }\epsilon >0\mathrm{\exists }\delta >0\mathrm{\forall }x\in D_f=\mathbb{R}\setminus \{1,2\},0<|x-1|<\delta :|\frac{x^3+x^2+3x+3}{x-2}+8|<\epsilon$$

Transform $\epsilon$ inequality

$$|\frac{x^3+x^2+3x+3}{x-2}+8|=\left|\frac{x^3+x^2+3x+3+8x-16}{x-2}\right|=$$$$=\left|\frac{x^3+x^2+11x-13}{x-2}\right|=|x-1|\cdot \frac{|x^2+2x+13|}{|x-2|}<\epsilon$$$$g(x)=\frac{|x^2+2x+13|}{|x-2|}$$

Assume that $0<|x-1|<\frac{1}{2}$.

$$-\frac{1}{2}<|x-1|<\frac{1}{2}⟹\frac{1}{2}<x<\frac{3}{2}$$

Then for denominator of $g(x)$

$$-\frac{3}{2}<x-2<-\frac{1}{2}⟹\frac{1}{2}<|x-2|<\frac{3}{2}⟹$$$$⟹\frac{2}{3}<\frac{1}{|x-2|}<2$$

Then for numerator of $g(x)$

$$x^2+2x+13<{\left(\frac{3}{2}\right)}^2+2\cdot \frac{3}{2}+13=\frac{9}{4}+3+13=18+\frac{1}{4}<19$$

So

$$|x-1|\cdot \frac{|x^2+2x+13|}{|x-2|}<|x-1|\cdot 19\cdot 2=38\cdot |x-1|$$$$38\cdot |x-1|<\epsilon ⟹|x-1|<\frac{\epsilon }{38}$$

So we can choose $\delta =min\{\frac{1}{2},\frac{\epsilon }{38}\}$.

$$\mathrm{\forall }\epsilon >0\mathrm{\exists }\delta =min\{\frac{1}{2},\frac{\epsilon }{38}\}\mathrm{\forall }x\in D_f=\mathbb{R}\setminus \{1,2\},$$$$0<|x-1|<\delta :|\frac{x^3+x^2+3x+3}{x-2}+8|<38\cdot |x-1|<\epsilon$$