• retagged by
1,134 views

3 Answers

Best answer
3 3 votes
RHL=$\lim_{x\rightarrow 0^{+}} sin(1/x)$   put $x\rightarrow 0+h$   then, RHL= $\lim_{h\rightarrow 0} sin(1/h)$

it gives positive value {-1 to 1}

 

LHL = $\lim_{x\rightarrow 0^{-}} sin(1/x)$  put $x\rightarrow 0-h$ then ,LHL=$\lim_{h\rightarrow 0} sin(-1/h) = \lim_{h\rightarrow 0} -sin(1/h)$ it gives negative value [-1 to 1]

so LHL !=RHL so limit doesnot exist
–1 –1 vote
(d) none of these

finite/0=infinity

so, 1/0=infinity

and -1<=sinx<=1

so, -1<=sin(infinity)<=1

so, none of these
Position:
Show:

No related questions found