What-are-the-conditions-whereby-the-limit-of-a-function-does-not-exist-at-a-poont-

Question Number 27727 by NECx last updated on 13/Jan/18

W h a t a r e t h e c o n d i t i o n s w h e r e b y

t h e l i m i t o f a f u n c t i o n d o e s n o t

e x i s t a t a p o o n t ?

Answered by prakash jain last updated on 13/Jan/18

When LHL is not equal to RHL .

Commented by NECx last updated on 13/Jan/18

i n t h i s c a s e \lim_{x \to 0} \frac{x^{2} + 3 x - 4}{2 - \sqrt{x^{2} + 4}}

h o w d o I f i n d t h e L H L a n d R H L

t o s h o w t h a t t h e l i m i t d o e s n o t

e x i s t i f a t a l l i t d o e s n t .

Commented by prakash jain last updated on 13/Jan/18

\frac{x^{2} + 3 x - 4}{2 - \sqrt{x^{2} + 4}}

in this 2 - \sqrt{x^{2} + 4} < 0 for all x

x^{2} + 3 x - 4 = - 4 near x = 0

\lim_{x \to 0} \frac{x^{2} + 3 x - 4}{2 - \sqrt{x^{2} + 4}}

= \lim_{x \to 0} \frac{- 4}{2 - \sqrt{x^{2} + 4}}

2 - \sqrt{x^{2} + 4} = 2 - {(4 + x^{2})}^{1 / 2} = 2 - 2 {(1 + \frac{x^{2}}{4})}^{1 / 2}

= 2 - 2 (1 + \frac{1}{2} \frac{x^{2}}{4} + \dots) = - (\frac{x^{2}}{8} + a l l + v e yerms)

hence limits exists and is infinity .

Commented by prakash jain last updated on 13/Jan/18

both LHL and RHL tend to +infinity