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


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

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

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


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


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


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

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

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

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



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

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


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

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

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



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


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

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


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


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


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


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


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


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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