# Proofs of all Limit Formulas

Some important limit formulas will be discussed here. The concept of the limit of a function is very useful in the theory of Calculus. In this post, we will prove all the important limit formulas one by one.

#### Trigonometric Functions Limit Formulas

At first, we will show that the limit of sin(x)/x is 1 when x tends to 0.

 Formula 1: $\, \lim\limits_{x \to 0} \dfrac{\sin x}{x}=1$

Brief Proof:

The proof is without applying L’Hospital’s rule.

It is known that

$\sin x \leq x \leq \tan x$, for all real x.

$\Rightarrow 1 \leq \dfrac{x}{\sin x} \leq \dfrac{\tan x}{\sin x}$

$\Rightarrow 1 \leq \dfrac{x}{\sin x} \leq \dfrac{1}{\cos x}$

Taking x tends to 0 on both sides, we get that

$\lim\limits_{x \to 0} 1$ $\leq \lim\limits_{x \to 0} \dfrac{x}{\sin x}$ $\leq\lim\limits_{x \to 0} \dfrac{1}{\cos x}$

$\Rightarrow 1 \leq \lim\limits_{x \to 0}\dfrac{x}{\sin x}\leq 1$

Thus, by the squeeze theorem on limits, we obtain that

$\lim\limits_{x \to 0} \dfrac{x}{\sin x}=1$

This implies that

$\lim\limits_{x \to 0} \dfrac{\sin x}{x}=1$

Also Read: Proof of all Limit Properties

#### Polynomial Functions Limit Formulas

Next, we prove that the limit of (xn-an)/(x-a) is nan-1 when x approaches a.

 Formula 2: $\, \lim\limits_{x \to a} \dfrac{x^n-a^n}{x-a}=na^{n-1}$

Proof:

Step 1: First, suppose that $n$ is a positive integer. Note that

$x^n-a^n=(x-a) \times$ $(x^{n-1}+x^{n-2}a+\cdots+a^{n-1})$

Using this fact, we have

$\lim\limits_{x \to a}\dfrac{x^n-a^n}{x-a}$

$=\lim\limits_{x\to a}(x^{n-1}+x^{n-2}a+\cdots+a^{n-1})$

$=a^{n-1}+a^{n-2}a+\cdots+a^{n-1}$

$=a^{n-1}+a^{n-1}a+\cdot\cdot$ till n-terms

$=na^{n-1}$

So the above formula holds for a positive integer $n.$

Step 2: Now, assume that $n$ is a negative integer. So write $n=-m$ for some positive integer $m.$

Now,  $\dfrac{x^n-a^n}{x-a}$ $=\dfrac{x^{-m}-a^{-m}}{x-a}$ $=\dfrac{\frac{1}{x^m}-\frac{1}{a^m}}{x-a}$ $=\dfrac{a^m-x^m}{a^mx^m(x-a)}$ $=-\dfrac{x^m-a^m}{a^mx^m(x-a)}$ $=-\dfrac{1}{a^mx^m}(x^{m-1}+x^{m-2}a+\cdots$ $+a^{m-1})$

$\therefore \lim\limits_{x \to a}\dfrac{x^n-a^n}{x-a}$

$=\lim\limits_{x \to a}\frac{-1}{a^mx^m}(x^{m-1}+x^{m-2}a^2+\cdots$ $+a^{m-1})$

$=\frac{-1}{a^ma^m}(a^{m-1}+a^{m-2}a+\cdots$  $\text{till m-terms})$

$=-\frac{1}{a^{2m}} \times ma^{m-1}$

$=-ma^{m-1-2m}$

$=-ma^{-m-1}$

$=na^{n-1} \quad$ $[\because n=-m]$

Hence, the formula is also true for a negative integer $n.$

Step 3: So we now assume that $n$ is a rational number. Write $n=\dfrac{p}{q},$ where $q \neq 1$ is a positive integer and $p$ is either positive or negative integer.

Let $x^{1/q}=z$ and $a^{1/q}=b.$

So $x=z^q$ and $a=b^q$

Note that $z \to b$ as $x \to a$

Now, $\dfrac{x^n-a^n}{x-a}$ $=\dfrac{x^{\frac{p}{q}}-a^{\frac{p}{q}}}{x-a}$ $=\dfrac{z^p-b^p}{z^q-b^q}$ $=\Big(\dfrac{z^p-b^p}{z-b} \Big)/\Big(\dfrac{z^q-b^q}{z-b} \Big)$

$\therefore \lim\limits_{x \to a}\dfrac{x^n-a^n}{x-a}$

$=\lim\limits_{x \to a} \Big(\dfrac{z^p-b^p}{z-b} \Big)/ \Big(\dfrac{z^q-b^q}{z-b} \Big)$

$=\lim\limits_{z \to b}\Big (\dfrac{z^p-b^p}{z-b} \Big)/\lim\limits_{z \to b} \Big(\dfrac{z^q-b^q}{z-b} \Big)$

$=pb^{p-1}/qb^{q-1}$

$=\frac{p}{q}b^{p-q}$

$=\frac{p}{q}a^{\frac{p-q}{q}}$ $[\because b=a^{1/q}]$

$=\frac{p}{q}a^{\frac{p}{q}-1}$

$=na^{n-1}$ $[\because n=\frac{p}{q}]$

Thus, for any rational number $n,$ we have

$\lim\limits_{x \to a} \dfrac{x^n-a^n}{x-a}=na^{n-1}$

As an application of the above formula, we show that ((1+x)n-1)/x is n when x tends to 0.

 Formula 3: $\, \lim\limits_{x \to 0}\dfrac{(1+x)^n-1}{x}=n$

Proof:

Let $1+x=z$.

So $z \to 1$ as $x \to 0$

Note that $x=z-1$

Now, $\lim\limits_{x \to 0}\dfrac{(1+x)^n-1}{x}$

$=\lim\limits_{z \to 1} \dfrac{z^n-1}{z-1}$

$=n \cdot 1^{n-1}$, by Formula 2

$=n$ ♣

#### Exponential Functions Limit Formulas

Now, we will prove that the limit of (ex-1)/x is 1 when x tends to 0. The proof will be based on Bernoulli’s Inequality and without using L’Hospital’s rule.

 Formula 4: $\, \lim\limits_{x \to 0} \dfrac{e^x-1}{x}=1$

Proof:

By Bernoulli’s Inequality, we have

$1+x\leq (1+\frac{x}{n})^n$ for $|x| \leq n$

Letting n tends to $\infty$, we get

$1+x\leq \lim\limits_{n \to \infty}(1+\frac{x}{n})^n$

$\Rightarrow 1+x \leq e^x$

[$\because \lim\limits_{n \to \infty}(1+\frac{x}{n})^n=e^x$]

$\Rightarrow x \leq e^x-1 \quad \cdots (1)$

Now, we know that

$e^{-x}=1-x+\frac{x^2}{2!}- \cdots$

Thus, for $|x| \leq 1$

$1-x \leq e^{-x}$

$\Rightarrow \frac{1}{1-x} \geq e^x$

$\Rightarrow \frac{1}{1-x}-1 \geq e^x-1$

$\Rightarrow \frac{x}{1-x} \geq e^x-1 \quad\cdots (2)$

Combining (1) and (2) we get for $|x|\leq 1$ that

$x \leq e^x-1 \leq \frac{x}{1-x}$

$\Rightarrow 1 \leq \frac{e^x-1}{x} \leq \frac{1}{1-x}$

Taking limit $x \to 0$ we obtain that

$1 \leq \lim\limits_{x \to 0} \frac{e^x-1}{x} \leq 1$

Thus, by squeeze theorem, we deduce that

$\lim\limits_{x \to 0} \frac{e^x-1}{x}=1$

Now, we show that the limit of (ax-1)/x is logea when x tends to zero.

 Formula 5: $\, \lim\limits_{x \to 0} \dfrac{a^x-1}{x}=\log_e a (a>0)$

Proof:

Let $a^x=e^z$

Taking logarithm on both sides, we have $x\log_e a=z$

Note that $z \to 0$ as $x \to 0$

Now, $\lim\limits_{x \to 0} \dfrac{a^x-1}{x}$

=$\lim\limits_{z \to 0} \Big(\dfrac{e^z-1}{z}/\dfrac{z}{\log_e a}\Big)$

=$\log_e a \cdot \lim\limits_{z \to 0}\dfrac{e^z-1}{z}$

=$\log_e a \cdot 1$, by Formula 4

=$\log_e a$ ♣

#### Logarithmic Functions Limit Formulas

Next, we will prove that the limit of loge(1+x)/x is 1 when x approaches 0.

 Formula 6: $\, \lim\limits_{x \to 0} \dfrac{\log_e(1+x)}{x}=1$

Proof:

Let $\log_e(1+x)=z$

$\therefore 1+x=e^z$

So $z \to 0$ as $x \to 0$

Now, $\lim\limits_{x \to 0} \dfrac{\log_e(1+x)}{x}$

=$\lim\limits_{z \to 0} \dfrac{z}{e^z-1}$

=$\large{\frac{1}{\lim\limits_{z \to 0} \frac{z}{e^z-1}} }$

=$\frac{1}{1}$,by Formula 3

=$1$ ♣

#### Application of Exponential or Logarithmic Limit Formulas

Next, we prove that the limit of (1+x)1/x is e as x tends to 0.

 Formula 7: $\, \lim\limits_{x \to 0}(1+x)^{\frac{1}{x}}=e$

Proof:

The above formula 6 can be written as

$\lim\limits_{x \to 0} \frac{1}{x}\log_e(1+x)=1$

$\Rightarrow \lim\limits_{x \to 0}\log_e(1+x)^{\frac{1}{x}}=1$

$\Rightarrow \log_e \big(\lim\limits_{x \to 0}(1+x)^{\frac{1}{x}}\big)=1$

$\Rightarrow \lim\limits_{x \to 0}(1+x)^{\frac{1}{x}}=e^1=e$

Next, we prove that the limit of (1+1/x)x is e as x tends to 0.

 Formula 8: $\, \lim\limits_{x \to \infty}(1+\frac{1}{x})^{x}=e$

Proof:

Let $\frac{1}{x}=z$

Then $z \to 0$ as $x \to \infty$

$\therefore \lim\limits_{x \to \infty}(1+\frac{1}{x})^{x}$

$=\lim\limits_{z \to 0}(1+z)^{\frac{1}{z}}$

$=e$, by Formula 7

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