proofs of various limit properties

To make the notation clearer, we let \[f(x)=c\] to get \[\lim_{x\to a}f(x)=c\]. Let \[\epsilon>0\] and we need to show that we can find \[\delta>0\] so that \[\left| f(x)-c \right|<\epsilon\] whenever \[0<\left| x-a \right|<\delta\].

Since we've defined \[f(x)=c\], \[\left| f(x)-c \right|<\epsilon\] is trivially satisfied as \[\epsilon\ne0\]. So choosing any \[\delta>0\], then \[\left| f(x)-c \right|=\left| c-c \right|=0<\epsilon\].

First we'll be looking at the case where \[c=0\]. \[\lim_{x\to a}0f(x)=\lim_{x\to a}0=0\]. Since multiplying by zero collapses all values to zero, then \[0=0\cdot f(x)\], thus proving the equation when \[c=0\].

Now, for the general case where \[c\ne0\]. Let \[\epsilon>0\], and since \[\lim_{x\to a}f(x)=K\] is true, then by definition there must exist a \[\delta_{1}>0\] such that \[\left| f(x)-K \right|<\frac{\epsilon}{\left| c \right|}\] whenever \[0<\left| x-a \right|<\delta_{1}\]. This is because the phrase in the definition of limit "for every \[\epsilon>0\]" tells us that you can choose any number as long as it's positive.

Since ultimately we want to show that \[\left| cf(x)-cK \right|<\epsilon\] whenever \[0<\left| x-a \right|<\delta\], we choose \[\delta=\delta_{1}\] to ensure that \[\left| f(x)-K \right|<\frac{\epsilon}{\left| c \right|}\] holds.
Then, assuming \[0<\left| x-a \right|<\delta\], \[\left| cf(x)-cK \right|=\left| c \right|\cdot \left| f(x)-K \right|\], which now we can write \[\left| c \right|\cdot \left| f(x)-K \right|<\left| c \right|\cdot \frac{\epsilon}{\left| c \right|}\], thus showing that \[\left| cf(x)-cK \right|<\epsilon\].

We'll start by proving \[\lim_{x\to a}[f(x)+g(x)]=\lim_{x\to a}f(x)+\lim_{x\to a}g(x)\]. To prove this, it is imperative to utilise \[\left| a+b \right|\le \left| a \right|+\left| b \right|\] (triangle inequality).

Let \[\epsilon>0\], \[\lim_{x\to a}f(x)=K\] and \[\lim_{x\to a}g(x)=L\], by definition there must be a \[\delta_{1}>0\] and \[\delta_{2}>0\] such that:

• \[\left| f(x)-K \right|<\frac{\epsilon}{2}\] whenever \[0<\left| x-a \right|<\delta_{1}\]
  
• \[\left| g(x)-L \right|<\frac{\epsilon}{2}\] whenever \[0<\left| x-a \right|<\delta_{2}\]
  

Now choose \[\delta=\min \left\{ \delta_{1},\delta_{2} \right\}\] (to ensure both inequalities containing \[\delta_{1}\] and \[\delta_{2}\] are true). Then, we need to show that \[\left| (f(x)+g(x))-(K+L) \right|<\epsilon\] whenever \[0<\left| x-a \right|<\delta\]. Assuming \[0<\left| x-a \right|<\delta\]:

\begin{align*} \left| (f(x)+g(x))-(K+L) \right|&=\left| (f(x)-K)+(g(x)-L) \right|\\ &\le \left| f(x)-K \right|+\left| g(x)-L \right|\text{by triangle inequality}\\ \end{align*}

Next, we substitute \[\frac{\epsilon}{2}\], which works due to our definition above:

\begin{align*} \left| (f(x)+g(x))-(K+L) \right|&< \frac{\epsilon}{2}+\frac{\epsilon}{2}\\ &<\epsilon \end{align*}

Do note that we've switched the sign from \[\le\] to \[<\], as \[\frac{\epsilon}{2}>\] both \[\left| f(x)-K \right|\] and \[\left| g(x)-L \right|\], thus \[\epsilon\] is impossible to be equal with \[\left| (f(x)+g(x))-(K+L) \right|\].

Thus we've proven that \[\lim_{x\to a}[f(x)+g(x)]=\lim_{x\to a}f(x)+\lim_{x\to a}g(x)\].

Next we'll prove \[\lim_{x\to a}[f(x)-g(x)]=\lim_{x\to a}f(x)-\lim_{x\to a}g(x)\] by utilising what we've proven previously.

\begin{align*} \lim_{x\to a}[f(x)-g(x)]&=\lim_{x\to a}[f(x)+(-1)g(x)]\\ &=\lim_{x\to a}f(x)+\lim_{x\to a}(-1)g(x)\\ &=\lim_{x\to a}f(x)+(-1)\lim_{x\to a}g(x)\\ &=\lim_{x\to a}f(x)-\lim_{x\to a}g(x)\\ \end{align*}

First we'll prove that \[\lim_{x\to a}[(f(x)-K)(g(x)-L)]=0\].

We'll separate that equation into two parts, \[\lim_{x\to a}[f(x)-K]=0\] and \[\lim_{x\to a}[g(x)-L]=0\]. Let \[\epsilon>0\], by definition there must exist a \[\delta_{1}>0\] and \[\delta_{2}>0\] such that:

• \[\left| (f(x)-K)-0 \right|<\sqrt{\epsilon}\] whenever \[0<\left| x-a \right|<\delta_{1}\]
  
• \[\left| (g(x)-L)-0 \right|<\sqrt{\epsilon}\] whenever \[0<\left| x-a \right|<\delta_{2}\]
  

Then choose \[\delta=\min \left\{ \delta_{1},\delta_{2} \right\}\]. Assuming \[0<\left| x-a \right|<\delta\] we then get:

\begin{align*} \left| (f(x)-K)(g(x)-L)-0 \right|&=\left| f(x)-K \right|\left| g(x)-L \right|\\ &<\sqrt{\epsilon}\sqrt{\epsilon}\\ &=\epsilon \end{align*}

Which proves that \[\lim_{x\to a}[(f(x)-K)(g(x)-L)]=0\].

Now, the reason why we chose to prove \[\lim_{x\to a}[(f(x)-K)(g(x)-L)]=0\] first is that \[(f(x)-K)(g(x)-L)=f(x)g(x)-Lf(x)-Kg(x)+KL\].

\begin{align*} f(x)g(x)&=(f(x)-K)(g(x)-L)+Lf(x)+Kg(x)-KL\\ \lim_{x\to a}[f(x)g(x)]&=\lim_{x\to a}[(f(x)-K)(g(x)-L)+Lf(x)+Kg(x)-KL]\\ &=\lim_{x\to a}[(f(x)-K)(g(x)-L)]+\lim_{x\to a}Lf(x)+\lim_{x\to a}Kg(x)-\lim_{x\to a}KL\\ &=0+L\lim_{x\to a}f(x)+K\lim_{x\to a}g(x)-\lim_{x\to a}KL\\ &=LK+KL-KL\\ &=LK\\ &=\lim_{x\to a}f(x)\cdot\lim_{x\to a}g(x)\\ \end{align*}

The reason why \[\lim_{x\to a}KL=KL\] is that \[KL=\lim_{x\to a}f(x)\cdot\lim_{x\to a}g(x)\], which makes \[KL\] a constant, and based on what we've proved above, for any constant \[c\], \[\lim_{x\to a}c=c\].

We'll start by proving \[\lim_{x\to a}\frac{1}{g(x)}=\frac{1}{\displaystyle\lim_{x\to a}g(x)}\]. Now, let \[\lim_{x\to a}g(x)=L\], by definition there must be a \[\delta_{1}>0\] such that \[\left| g(x)-L \right|<\frac{\left| L \right|}{2}\] whenever \[0<\left| x-a \right|<\delta_{1}\]. Assuming \[0<\left| x-a \right|<\delta_{1}\]:

\begin{align*} \left| L \right|&=\left| L-g(x)+g(x) \right|\\ &\le \left| L-g(x) \right|+\left| g(x) \right|\text{by triangle inequality}\\ &\le \left| g(x)-L \right|+\left| g(x) \right|\\ &<\frac{\left| L \right|}{2}+\left| g(x) \right|\\ \end{align*}

Moving the \[\frac{\left| L \right|}{2}\] to the left-hand side gives us \[\frac{\left| L \right|}{2}<\left| g(x) \right|\implies \frac{1}{g(x)}<\frac{2}{\left| L \right|}\].

Let \[\epsilon>0\]. By definition there must be also a \[\delta_{2}>0\] such that \[\left| g(x)-L \right|<\frac{\left| L \right|^{2}}{2}\epsilon\] whenever \[0<\left| x-a \right|<\delta_{2}\]. Then, we choose our \[\delta=\min \left\{ \delta_{1},\delta_{2} \right\}\]. Assuming \[0<\left| x-a \right|<\delta\]:

\begin{align*} \left| \frac{1}{g(x)}-\frac{1}{\left| L \right|} \right|&=\left| \frac{L-g(x)}{Lg(x)} \right|\\ &=\frac{1}{\left| Lg(x) \right|}\left| L-g(x) \right|\\ &=\frac{1}{\left| L \right|}\frac{1}{\left| g(x) \right|}\left| g(x)-L \right|\\ &<\frac{1}{\left| L \right|}\frac{2}{\left| L \right|}\left| g(x)-L \right|\\ &<\frac{2}{\left| L \right|^{2}}\frac{\left| L \right|^{2}}{2}\epsilon\\ &<\epsilon\\ \end{align*}

Since we've proven that \[\lim_{x\to a}\frac{1}{g(x)}=\frac{1}{\displaystyle\lim_{x\to a}g(x)}\],

\begin{align*} \lim_{x\to a}\left[ \frac{f(x)}{g(x)} \right]&=\lim_{x\to a}f(x)\cdot\lim_{x\to a}\frac{1}{g(x)}\\ &=\lim_{x\to a}f(x)\cdot \frac{1}{\displaystyle\lim_{x\to a}g(x)}\\ &=\frac{\displaystyle\lim_{x\to a}f(x)}{\displaystyle\lim_{x\to a}g(x)}\\ \end{align*}

For this case we'll be using proof by induction.

Assume \[n=1\], \[\lim_{x\to a}\left[ f(x) \right]^{1}=\left[ \lim_{x\to a}f(x) \right]^{1}\].
Assume \[n=2\], \[\lim_{x\to a}\left[ f(x) \right]^{2}=\lim_{x\to a}f(x)\cdot\lim_{x\to a}f(x)=\left[ \lim_{x\to a}f(x) \right]^{2}\]. Thus it is proven to be true.

Now on to our general case. Assume \[n=k\], \[n\in\mathbb{R}\] and \[\left[ f(x) \right]^{k+1}=\left[ f(x) \right]^{k}\cdot f(x)\].

\begin{align*} \lim_{x\to a}\left[ f(x) \right]^{k+1}&=\lim_{x\to a}\left( \left[ f(x) \right]^{k}\cdot f(x) \right)\\ &=\lim_{x\to a}\left[ f(x) \right]^{k}\cdot\lim_{x\to a}f(x)\\ &=K^{k}\cdot K\\ &=K^{k+1}\\ &=\left[ \lim_{x\to a}f(x) \right]^{k+1} \end{align*}

Let \[p(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}\] is a polynomial of degree \[n\],

\begin{align*} \lim_{x\to \infty}p(x)&=\lim_{x\to\infty}\left( a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0} \right)\\ &=\lim_{x\to\infty}a_{n}x^{n}\cdot\lim_{x\to\infty}\left( 1+\frac{a_{n-1}}{a_{n}x}+\cdots+\frac{a_{1}}{a_{n}x^{n-1}}+\frac{a_{0}}{a_{n}x^{n}} \right)\\ &=\lim_{x\to\infty}a_{n}x^{n}\cdot\lim_{x\to\infty}1\\ &=\lim_{x\to\infty}a_{n}x^{n}\\ \end{align*}

The reason the latter limit disappears is that when \[x\] approaches both \[\infty\] and \[-\infty\], the fractions will all be reduced to 0.

Let \[\epsilon>0\], we need to show that there is a \[\delta>0\] such that \[\left| f(g(x))-f(b) \right|<\epsilon\] whenever \[0< \left| x-a \right|<\delta\].

Assume \[f(x)\] is continuous at \[x=b\], then \[\lim_{x\to b}f(x)=f(b)\] so there must be a \[\delta_{1}>0\] such that \[\left| f(x)-f(b) \right|<\epsilon\] whenever \[0<\left| x-b \right|<\delta_{1}\].

Since, \[\lim_{x\to a}g(x)=b\], this indicates that there must be a \[\delta>0\] such that \[\left| g(x)-b \right|<\delta_{1}\] whenever \[0<\left| x-a \right|<\delta\] (the \[\delta_{1}\] is just a sort of replacement for \[\epsilon\]).

So, if \[g(x)\] is close to \[b\], then \[f(g(x))\] is close to \[f(b)\] as \[f\] is continuous at \[b\]. Thus, plugging the values in we get \[\left| f(g(x))-f(b) \right|<\epsilon\], proving that \[\lim_{x\to a}f(g(x))=f \left( \lim_{x\to a}g(x) \right)\].