SHORT NOTES: Limits , Continuity and Differentiability

Limits , Continuity and Differentiability

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A limit is the value that a function or sequence "approaches" as the input or index approaches some value

Suppose

f

is a real-valued function and

c

is a real number. The expression

$\lim_{x \to c}f(x) = L$

means that

f(x)

can be made to be as close to

L

as desired by making

x

sufficiently close to

c

. In that case, the above equation can be read as "the limit of

f

x

, as

x

approaches

c

, is

L

For example

Left hand and right hand limit of a function

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

$x$ can ‘approach’ or tends to '0' in two ways, either from the left hand side or from the right hand side:

$x \to {0^ - }$ :approach is from left side of $0$

$x \to {0^ + }$ : approach is from right side of $0$

e.g. $\lim_{x\rightarrow 0-}Sgn \ x=-1$ $\lim_{x\rightarrow 0+}Sgn \ x=1$

Sgn is signum function

The limit of $f(x)$ at $x = a$ is said to exist if the function approaches the same value from both sides i.e.

$\lim_{x\rightarrow a-}f(x)=\lim_{x\rightarrow a+}f(x)=l$

$\lim_{x\rightarrow a}f(x)=l$

Limits of the form $\dfrac{0}{0},\dfrac{\infty }{\infty },\infty - \infty ,\infty \times 0,{1^\infty },{0^\infty }$ are called indeterminate forms(limit may or mayn't exist). Such forms need to be reduced into determinate forms for which the limit can be determined.

Algebra of Limits

Let $\mathop {\lim }\limits_{x \to a} f\left( x \right) = l$ and $\mathop {\lim }\limits_{x \to a} g\left( x \right) = m.$

$\mathop {\lim }\limits_{x \to a} \left\{ {f\left( x \right) \pm g\left( x \right)} \right\} = \left\{ {\mathop {\lim }\limits_{x \to a} f\left( x \right)} \right\} \pm \left\{ {\mathop {\lim }\limits_{x \to a} g\left( x \right)} \right\} = l \pm m$
$\mathop {\lim }\limits_{x \to a} \left\{ {f\left( x \right) \cdot g\left( x \right)} \right\} = \left\{ {\mathop {\lim }\limits_{x \to a} f\left( x \right)} \right\} \cdot \left\{ {\mathop {\lim }\limits_{x \to a} g\left( x \right)} \right\} = l \cdot m$

$\mathop {\lim }\limits_{x \to a} \left\{ {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right\} = \dfrac{{\left\{ {\mathop {\lim }\limits_{x \to a} f\left( x \right)} \right\}}}{{\left\{ {\mathop {\lim }\limits_{x \to a} g\left( x \right)} \right\}}} = \dfrac{l}{m}{\rm{if}}m \ne 0$
$\mathop {\lim }\limits_{x \to a} \left\{ {Kf\left( x \right)} \right\} = K\left\{ {\mathop {\lim }\limits_{x \to a} f\left( x \right)} \right\} = Kl$
$\mathop {\lim }\limits_{x \to a} {\left( {f\left( x \right)} \right)^{g\left( x \right)}} = {\left( {\mathop {\lim }\limits_{x \to a} f\left( x \right)} \right)^{\left( {\mathop {\lim }\limits_{x \to a} g\left( x \right)} \right)}} = {l^m},$
$\mathop {\lim }\limits_{x \to a} f\left( {g\left( x \right)} \right) = f\left( {\mathop {\lim }\limits_{x \to a} g\left( x \right)} \right) = f\left( m \right)$
$\mathop {\lim }\limits_{x \to a} \left| {f\left( x \right)} \right| = \left| {\mathop {\lim }\limits_{x \to a} f\left( x \right)} \right| = \left| l \right|$
$\lim_{x\rightarrow a}f(x)=\infty$ then $\lim_{x\rightarrow a}\frac{1}{f(x)}=0$

Sandwich theorem:

If $\mathop {\lim }\limits_{x \to a} g\left( x \right) = \mathop {\lim }\limits_{x \to a} h\left( x \right) = l,$ and a function f(x) is such that

$g\left( x \right) \le f\left( x \right) \le h\left( x \right)$

for all x in neighborhood of a then

$\mathop {\lim }\limits_{x \to a} f\left( x \right) = l$

Existence of limit at $x = a$ implies $LHL = RHL$ Continuity of $f(x)$ at $x = a$ implies $LHL = RHL = f(a)$

Methods for Evaluation of Limits

(A) DIRECT SUBSTITUTION:

$\lim_{x\rightarrow 0}\frac{\cos x}{1+\sin x}=\frac{\cos 0}{1+\sin 0}=1$

(B) FACTORIZATION:

$\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^2} - 3x + 2}}{{{x^2} - 4}} = \mathop {\lim }\limits_{x \to 2} \dfrac{{(x - 1)(x - 2)}}{{(x - 1)(x + 2)}} = \mathop {\lim }\limits_{x \to 2} \dfrac{{x - 1}}{{x + 2}} = \dfrac{1}{4}$ $\left ( \frac{0}{0} \right )$

(C) RATIONALIZATION:

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {{x^2} + 1} - 1}}{{\sqrt {{x^2} + 16} - 4}}\left( {{\rm{of\, \,the\, \,indeterminate\,\, form \,\,}}\dfrac{{\rm{0}}}{{\rm{0}}}} \right)$

$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {{x^2} + 1} - 1}}{{\sqrt {{x^2} + 16} - 4}} \times \dfrac{{\sqrt {{x^2} + 1} + 1}}{{\sqrt {{x^2} + 1} + 4}} \times \dfrac{{\sqrt {{x^2} + 16} + 4}}{{\sqrt {{x^2} + 16} + 4}}$

$\mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2}}}{{{x^2}}} \times \dfrac{{\sqrt {{x^2} + 16} + 4}}{1}$

$= \dfrac{8}{2} = 4$

(D) REDUCTION TO STANDARD FORMS:

(i) $\lim_{x\rightarrow a} \frac{x^{n}-a}{x-a}=na^{n-1}$

(ii) $\lim_{x\rightarrow 0} \left ( 1+x \right )^\frac{1}{x}=e$ $\lim_{x\rightarrow \infty } \left ( 1+\frac{1}{x} \right )^x=e$

(iii) $\lim_{x\rightarrow 0 } \left\left ( 1+\alpha x \right )^\frac{\beta }{x}=e^{\alpha \beta }$ $\lim_{x\rightarrow \infty } \left ( 1+\frac{\alpha }{x} \right )^\beta x=e^{\alpha \beta }$

(iv) $\lim_{x\rightarrow 0 } \left\left ( 1+ x \right )^{\frac{1}{x}+\alpha }=e$ $\lim_{x\rightarrow \infty } \left\left ( 1+ \frac{1}{x} \right )^{{x}+\alpha }=e$

(v) $\lim_{x\rightarrow 0 } \frac{\log_{a}{(1+\alpha x)}}{x}=\alpha \log_{a}e$ , $\lim_{x\rightarrow 0 } \frac{\ln{(1+\alpha x)}}{x}=\alpha$

(vi) $\lim_{x\rightarrow 0 }\frac{a^{\alpha x}-1}{x}=\alpha \ln a$ , $\lim_{x\rightarrow 0 }\frac{e^{\alpha x}-1}{x}=\alpha$

(vii) $\lim_{x\rightarrow 0 }\frac{\sin \alpha x}{x}=\alpha$ , $\lim_{x\rightarrow 0 }\frac{\sin \alpha x}{\sin \beta x}=\frac{\alpha}{\beta }$

(viii) $\lim_{x\rightarrow 0 }\cos x=1$

(ix) $\lim_{x\rightarrow 0 }\frac{\tan x}{x}=1$

(E)Using L'Hospital rule:

If f(x) and g(x) are continuous and differentiable at x=a,

& $\lim_{x\rightarrow a }f(x)=\lim_{x\rightarrow a }g(x)=0$ or $\infty$ then

$\lim_{x\rightarrow a }\frac{f(x)}{g(x)}=\lim_{x\rightarrow a }\frac{f'(x)}{g'(x)}$

we can repeat this process till limit is evaluated

(F)Evaluation of left hand and right hand limit:

RHL of

$f(x)=\left\{\begin{matrix} \frac{\left | x-4 \right |}{x-4} ,\ x\neq 4\\ 0, \ x=4 \end{matrix}\right.$ at x=4

A function f(x) is said to be continuous at x=a if

$\lim_{x\rightarrow a^-}f(x)=\lim_{x\rightarrow a^+}f(x)=f(a)$

i.e LHL=RHL=value of function at x=a

Differentiability of a function at a point:

f(x) ,defined in (a,b), is said to be differentiable at x=c where $c\in (a,b)$

iff $\lim_{x\rightarrow c}\frac{f(x)-f(c)}{(x-c)}$ exists finitely

i.e.
$\lim_{x\rightarrow c^-}\frac{f(x)-f(c)}{(x-c)}=\lim_{x\rightarrow c^+}\frac{f(x)-f(c)}{(x-c)}$
$\Rightarrow \lim_{h\rightarrow 0}\frac{f(c-h)-f(c)}{-h}=\lim_{h\rightarrow 0}\frac{f(c+h)-f(c)}{h}$

$\Rightarrow$ LEFT HAND DERIVATIVE = RIGHT HAND DERIVATIVE
Relationship between continuity and differentiability

If a function is differentiable at a point , it is continuous at that point.But the converse is not necessarily true.

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$\lim_{x\rightarrow 4^{+}}f(x)=\lim_{h\rightarrow 0}f(4+h)=\lim_{h\rightarrow 0}\frac{\left | 4+h-4 \right |}{4+h-4}=\lim_{h\rightarrow 0}\frac{\left | h \right |}{h}$

$\lim_{h\rightarrow 0} \frac{h}{h}=\lim_{h\rightarrow 0} 1=1$

similarly for LHL evaluation $x\rightarrow a^-$ is replaced by $h\rightarrow 0$ where $x=a-h$

Continuity of a function: