Introduction To Limits

INTRODUCTION

Consider the function$f\text{(x)}\text{=}\frac{{\text{x}}^{\text{2}}-1}{x-1}$. Clearly $f\left(x\right)$is not defined at x = 1. At x = 1, $f\text{(x)}\text{=}\frac{\text{1}-1}{1-1}=\frac{0}{0}$, which is meaningless.

From the above table it is clear that as x approaches to 1 i.e. x $\to$ 1 from the left hand side (means x approaches 1 from the values less than 1)$f\left(x\right)$ approaches to 2 i.e. $f\left(x\right)\to 2$ The number 2 is called the left limit of $f\left(x\right)$ and in symbol we shall write $\underset{\text{x}\to {\text{1}}^{\text{ - }}}{\lim }f\text{(x)}\text{ = }\text{2}$.

Again let us study the behaviour of f(x) where x approaches towards 1 from the right-hand side.

It is clear from the table that as x approaches to 2 i.e. x $\to$ 2, from the right-hand side (means x approaches 1 from the values greater than 1) f(x) approaches to 2 i.e.$f\left(x\right)\to 2$ Here 2 is called the right-hand limit of f(x) and in symbol we will write $\underset{\text{x}\to {\text{1}}^{\text{+}}}{\lim }f(x)=2$

Thus we see that f(x) is not defined at x = 1 but its left-hand limit and right-hand limit as x ® 1 exist and are equal. When $\underset{\text{x}\to {\text{1}}^{\text{-}}}{\lim }f(x)\text{and }\underset{\text{x}\to \text{1+}}{\lim }f(x)$ are equal we say $\underset{\text{x}\to {\text{1}}^{}}{\lim }f(x)$ exist and is equal to 2.

Meaning of (x $\to$ a)

Let x be a variable and 'a' be a constant. x assumes value nearer and nearer to 'a', then we say 'x tends to a' and write 'x $\to$ a' and it doesn't mean x = a.

CONCEPT OF LIMIT

Let y = f(x) be a given function defined in the neighbourhood of x = a, but not necessarily at the point x = a. The limiting behaviour of the function in the neighbour of x = a is called the limit of the function when x approaches a. Mathematically we write this as $\underset{\text{x}\to \text{a}}{\lim }f(x)$ .

$\underset{\text{x}\to \text{a}}{\lim }f(x)=l$ would simply mean that when we approach the point x = a from the value which are just greater than or just smaller than x = a, f (x) would have a tendency to move closer to the value ${}^{\prime }{l}^{\prime }$.

RIGHT AND LEFT HAND LIMIT

Right-hand limit means tendency of function when we approach x = a from the value which just greater than 'a' and we write$\underset{\text{x}\to {\text{a}}^{+}}{\lim }f(x)$ .

Working rule to evaluate $\underset{\text{x}\to {\text{a}}^{+}}{\lim }f(x)$

$\bullet$Put x = a + h in f(x) to get $\underset{h\to 0}{\lim }f(a+h)$

$\bullet$Take the limit as h $\to$ 0

Left-hand limit means tendency of function when we approach x = a from the values which are just less than 'a' and we write $\underset{\text{x}\to {\text{a}}^{-}}{\lim }f(x)$.

Working rule to evaluate $\underset{\text{x}\to {\text{a}}^{-}}{\lim }f(x)$

$\bullet$Put x = a – h in f(x) to get $\underset{\text{h}\to 0}{\lim }f(a-h)$

$\bullet$Take the limit as h $\to$ 0

For example: $\underset{\text{x}\to 1^{+}}{\lim }\frac{x^2-1}{x-1}=\underset{\text{h}\to 0}{\lim }\frac{{(1+h)}^2-1}{(1+h)-1}=\underset{\text{h}\to 0}{\lim }\frac{h^2-2h}{h}=\underset{\text{h}\to 0}{\lim }(h+2)=2$

Thus for the existence of the limit of f(x) at x=a, it is necessary and sufficient that $\underset{\text{x}\to {\text{a}}^{+}}{\lim }f(x)$= $\underset{\text{x}\to {\text{a}}^{-}}{\lim }f(x)$, if these are finite or $\underset{\text{x}\to {\text{a}}^{+}}{\lim }f(x)$and $\underset{\text{x}\to {\text{a}}^{-}}{\lim }f(x)$ both should be either +$\infty$ or -$\infty$.

REASON FOR NON-EXISTENCE OF THE LIMIT

Following are the reasons when $\underset{\text{x}\to a}{\lim }f(x)$ will not exist.

Reason 1: when left and right tendencies of f(x) are not same in the neighbourhood of $x=a\Rightarrow$ $\underset{\text{x}\to a^{-}}{\lim }f(x)\ne \underset{\text{x}\to a^{+}}{\lim }f(x)$

For example $\underset{\text{x}\to 2}{\lim }[x],$ (when [.] denotes the greatest integer function)

It is clear from the figure that left tendency of [x] at x = 2 is 1 while the right tendency of [x] at x = 2 is 2. This $\underset{\text{x}\to 2}{\lim }[x],$will not exist.

Reason 2: If f(x) is not defined in the neighbourhood of x = a. For example $\underset{\text{x}\to 0}{\lim }{\sec }^{-1}\left(\cos x\right)$Here f(x) is well defined at x = 0

But in the left neighbourhood of 0, means $\underset{x\to 0^{-}}{\lim }{\sec }^{-1}(\cos x)$ , when

$x\to 0^{-}\cos x<1\text{and}{\sec }^{-1}(\cos x)\text{doesn }{}^{\prime }\text{ t}\text{exist}$, similarly for right tendency $\underset{\text{x}\to 0^{+}}{\lim }{\sec }^{-1}\left(\cos x\right)$, in this case also in the right neighbourhood of x=0, cosx<1. Hence sec^-1(cosx) is not defined.

Reason 3: When f(x) doesn't have a unique tendency. For example $\underset{\text{x}\to \text{0}}{\lim }\sin \frac{1}{x}$ we have that –1 $⩽$ sinx $⩽$ 1, that means for all non-zero values of x, sin 1/x would assume finite values. But when x becomes very near to zero sin 1/x would erratically oscillate between –1 and + 1. It implies that sin 1/x wouldn't have unique tendency for very small value of x. Thus $\underset{\text{x}\to \text{0}}{\lim }\text{ sin 1/x }$will not exist.

BASIC THEOREMS ON LIMITS

Let $\underset{x\to a}{\lim }$f(x) = ${\ell }_1$ and $\underset{x\to a}{\lim }$ g(x) = ${\ell }_2$, where ${\ell }_1$ and ${\ell }_2$ are finite, then the following theorems on limits can be used to evaluate the limits

(i) $\underset{x\to a}{\lim }$ (c₁ f(x) $\pm$ c₂ g(x)) = c₁ ${\ell }_1\pm c_2{\ell }_2$, where c₁and c₂ are given constants.

(ii) $\underset{x\to a}{\lim }$f(x). g(x) = $\underset{x\to a}{\lim }$ f (x). $\underset{x\to a}{\lim }$g (x) = ${\ell }_1$ ${\ell }_2$

(iii) $\underset{x\to a}{\lim }$ $\frac{f(x)}{g(x)}=\frac{\underset{x\to a}{\lim }f(x)}{\underset{x\to a}{\lim }g(x)}=\frac{{\ell }_1}{{\ell }_2},{\ell }_2\ne 0$

(iv) $\underset{x\to a}{\lim }$ f (g(x)) = f ($\underset{x\to a}{\lim }$g(x)) = f$\left({\ell }_2\right)$, if and only if f(x) is continuous at x = ${\ell }_2$.

For example $\underset{x\to 0}{\lim }$$\left[\frac{\sin x}{x}\right]$ $\ne \left[\underset{x\to 0}{\lim }\frac{\sin x}{x}\right]$ (where [.]denotes the greatest integer function)Here [x] is not continuous at x = 1. Also $\underset{x\to 0}{\lim }$$\left[\frac{\sin x}{x}\right]$=0 and $\left[\underset{x\to 0}{\lim }\frac{\sin x}{x}\right]$=1.

(v) If f(x) $⩽$ g(x) $\forall$ x $\in$ R, then $\underset{x\to a}{\lim }$ f(x) $⩽$ $\underset{x\to a}{\lim }$ g(x).

Note: We have to be very careful while using these theorems. For example if we try to apply the theorems on $\underset{x\to 0}{\lim }$$\frac{\sin x}{x}$=1 we get $\underset{x\to 0}{\lim }$$\frac{\sin x}{x}$= $\underset{x\to 0}{\lim }$ sin x.

$\underset{x\to 0}{\lim }$

. 1/x, which does not exist.

Which is an absurd result, because in this case the given limit can not be written as the product of two limits as $\underset{x\to 0}{\lim }$ 1/x does not exist.