# Indeterminate Forms

• ### Standard Limits

${\lim \limits_{x \to 0} \frac {sin(x)}{x} = 1, \lim \limits_{x \to 0} \frac {tan(x)}{x} = 1, \lim \limits_{x \to 0} \frac {1 - cos(x)}{x^2} = \frac 1 2}$

${\lim \limits_{x \to a } \frac {x^n - a^n}{x-a} = n \cdot x^{n-1}, \lim \limits_{x \to 0} \frac {a^x -1 }{x} = ln(a), \lim \limits_{x \to 0} (1+x)^{\frac 1 x} = e}$

${\lim \limits_{x \to 0+} ln(x) = -\infty , \lim \limits_{x \to \infty} ln(x) = \infty}$

• ### Introduction

Consider 2 functions, ${f(x)}$ and ${g(x)}$. Let ${x=a}$ be a value at which these functions are defined.

I) If ${f(x) =sin (x), g(x) =x}$ and ${a=0}$, then the limit ${\lim \limits_{x \to 0} \frac {sin(x)}{x}}$ takes the form ${\frac 0 0}$.

II) If ${f(x)= \frac 1 x, g(x) = tan \Big( \frac \pi 2 - x\Big), a= 0}$, then ${\lim \limits_{x \to 0} \frac {\frac 1 x}{ tan \Big( \frac \pi 2 - x\Big)}}$ takes the form ${\frac {\infty}{\infty}}$

III) If ${f(x) = x, g(x) = ln(x)}$ and ${a=0}$, then the limit ${\lim \limits_{x \to 0} x \cdot ln(x)}$ takes the form ${0 \times \infty}$

IV) If ${f(x)= sin(x), g(x)=x}$, and ${a=0}$, then the limit ${\lim \limits_{x \to 0} [{sin(x)}]^x}$ takes the form ${0^0}$.

V) If ${f(x)= cos(x), g(x)=tan (\frac {\pi}{2} - x)}$, then the limit ${\lim \limits_{x \to 0} {[cos(x)]}^{tan (\frac \pi 2 -x)}}$ takes the form ${1^\infty}$.

VI) If ${f(x)= \frac 1 x}$ and ${g(x)= sin(x)}$ and ${a=0}$, then the limit ${\lim \limits_{x \to 0} \Big(\frac {1}{x} \Big)^{sin(x)}}$ takes the form ${\infty^0}$

VII) If ${f(x)= tan (\frac \pi 2 -x)}$ and ${g(x)= \frac 1 x}$ and ${x=0}$, then the limit ${\lim \limits_{x \to a} tan (\frac \pi 2 -x) - \frac 1 x}$ takes the form ${\infty - \infty}$.

These are known as the indeterminate forms. The limits are evaluated either by L’Hosptial’s rule or by substituting an equivalent infinitesimal.

• ### L’Hosptial’s Rule (French : ${Lopital}$)

The rule can be proved using Taylor’s theorem. It says, if ${f(x)}$ and ${g(x)}$ are ${0}$ at ${x = a}$ or ${\lim \limits_{x \to a} f(x)=0}$ and ${\lim \limits_{x \to a} g(x)=0}$, then

${\lim \limits_{x \to a } \frac {f(x)}{g(x)} = \lim \limits_{x \to a} \frac {f'(x)}{g'(x)}}$

This rule is sometimes applied on ${n}$th derivatives, if all derivatives of lesser orders are ${0}$.

• ### Equivalent Infinitesimal

This is used for evaluation of ${\frac 0 0}$ form. One of the functions can be replaced by another, if they both converge to ${0}$ at a point and the limit of their ratio at that point is ${1}$. For example,

${x \sim sin(x), x \sim tan^{-1}x, \sim ln(1+x)}$

• ### Hint for MCQs

One can try substituting a value of ${x}$ closer to (but not equal to) the actual limit. Evaluate the function using the calculator. The answer will be closer to the actual limit. We’ve actually used the concept of limit here.

Explanation: Consider the limit

${\lim \limits_{x \to 0} [1+ sin(x)]^{cot(x)}}$

This is of the form ${1^{\infty}}$. Let’s put ${x=0.1}$ in the function ${[1+ sin(x)]^{cot(x)}}$.

We get ${[1+sin(0.1)]^{cot(0.1)}}$ as ${2.58160}$.

On substituting ${x = 0.01}$, a value closer to ${0}$, we get ${2.7046}$.

On substituting ${x = 0.001}$, a value closer to ${0}$ and ${< 0.01}$, we get ${2.7169}$. Clearly, the limit is ${e}$.