# Limits

A number ‘l’ is called limit of a function f(x) when $x \to a$ i.e., $\overset{lim}{x \to a} f (x) = l$ if given $\epsilon > 0$ , there exists $\delta > 0$ such that |x –a| $< \delta \to$ | f(x) – l | < $\epsilon$ .

### Right hand and left hand limits

Let h be a small positive number. Left hand side limit of f(x) when $x \to a$ , is denoted by f(a -0) and is defined as: $F ( a - 0 ) = \underset{h \to 0}{lim} f (a - h)$

Right hand side limit of f(x), when $x \to a$ , is denoted by f(a + 0) and is defined as: $f ( a + 0) = \underset{h \to 0}{lim} f(a + h)$ $\underset{x \to a}{lim} f (x)$ exists if $\underset{h \to 0}{lim} f ( a + h ) = \underset{h \to 0}{lim} f ( a - H)$

### Indeterminate forms

If a function f(x) takes the form $f (x) = \dfrac{0}{0} \, \, at \, \, x = a$ , then say that f(x) is indeterminate at x=a. Other Indeterminate Forms are $\dfrac{ \infty}{ \infty} , \infty - \infty , 0^0 , \infty ^0 , 1^{ \infty} , 0 \times \infty$ .

### L’ hospital’s rule

If $\phi (x)$ and $\psi (x)$ are functions of x such that $\phi (a) = 0 = \psi (a)$ , then $\underset{x \to a}{lim} \dfrac{ \phi (x)}{ \phi (x)} = \underset{x \to a}{lim} \dfrac{\phi' (x)}{\psi' (x)}$

### The form $0 \times \infty$

This form can easily be reduced either to form $\dfrac{0}{0}$ of $\dfrac{ \infty}{ \infty}$ .

Example:

Evaluate $\underset{x \to 1}{lim} \, \, sec ( \dfrac{ \pi r}{2} ) log x$

Solution: $= \underset{x \to 1}{lim} \, \, \, sec ( \dfrac{ \pi x}{2} ) log x$ $= \underset{x \to 1}{lim} ( \dfrac{ log x}{ cos ( \pi x / 2} )$ $= \underset{x \to 1}{lim} \dfrac{1 / x}{- \dfrac{ \pi}{2} sin ( \dfrac{ \pi x}{2} )} = - \dfrac{2}{\pi}$

### The form $\infty - \infty$

This can also be reduced to the form $\dfrac{0}{0} \, or \, \dfrac{ \infty}{\infty}$

Example:

Evaluate $\underset{x \to 0}{lim} ( \dfrac{1}{x^2} - \dfrac{1}{sin ^2 x} )$ $= \underset{x \to 0}{lim} \dfrac{sin ^2 x - x^2}{x ^2 sin ^2 x}$ $= \underset{x \to 0}{lim} \dfrac{sin ^2 x - x^2}{x^4} ( \dfrac{1}{1} ) ^2$ $= \underset{x \to 0}{lim} \dfrac{ sin (2x) - 2x}{4x^3}$ $= \underset{x \to 0}{lim} \dfrac{ [ 2x - \dfrac{2x^3}{3 !} + \dfrac{ 2x ^5}{5 !} - \cdots ] - 2x}{4x^3}$ $= - \dfrac{8}{3 !} \dfrac{1}{4} = - \dfrac{1}{3}$

### Sandwich Theorem (or Squeeze principle)

If f, g, h are functions such that $f (x) \leq g (x) \leq h (x)$ for all x in the neighborhood of a and if $\underset{x \to a}{lim} f (x) = l$ , $\underset{x \to a}{lim} h (x) = l , \underset{x \to a}{lim} g (x) = l$

### Algebra of limits $if \, \, \, \underset{x \to a}{lim} f (x) = l$ , $\underset{x \to a} g (x) = m , \, \, then$ $( 1 ) \, \, \, \underset{x \to a}{lim} [ f (x) \pm g (x) ] = \underset{x \to a}{lim} f (x) \pm \underset{ x \to a}{lim} g (x) = l \pm m$ $(2) \, \, \, \underset{x \to a}{lim} f (x) g(x) = \underset{x \to a}{lim} f(x) \underset{x \to a }{lim} g (x) = lm$ $(3) \underset{x \to a}{lim} \dfrac{f (x)}{g (x)} = \dfrac{ \underset{x \to a}{ lim} f (x)}{ \underset{x \to a}{lim} g(x)} = \dfrac{l}{m}$ $(4) [ \underset{x \to a}{lim} f(x) ] ^n = l^n \, \, if \, \, n > 0$

### Evaluation of exponential limits of the form $1^{\infty}$

Result:

(i) If $\underset{x \to a}{lim} f (x) = \underset{x \to a}{lim} g (x) = 0 , \, \, then$ $= \underset{x \to a}{lim} [ 1 + f(x) ]^{1 /g(x)} = e \underset{x \to a}{lim} \dfrac{f(x)}{g(x)}$

(ii) If $\underset{x \to a}{lim} f(x) = 1 , \underset{x \to a}{lim} g (x) = \infty , then$ $= \underset{x \to a}{lim} [ f(x)^{g (x)} = \underset{x \to a}{lim} [ 1 + [ f(x) - 1]^{g(x)}$ $= e \, \, \underset{x \to a}{lim} [ f(x) -1 ] g(x)$

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