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$\lim_{x\rightarrow 0}\sin \left( \frac{1}{x}\right)$

(a) 1

(b)  0

(c) does not exist

(d) none of these
in Calculus by Active (5.1k points)
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3 Answers

+3 votes
Best answer
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
by Active (5.1k points)
+2 votes

(c) option

lt x--->0 {sin(1/x)/(1/X)}*(1/x)

blue term give 1 when after that we will apply limit to(1/x)

so lt x-->0(1/x)

which gives not defined...

by Boss (20.1k points)
edited by
0
how?
–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
by (205 points)
0
take RHL AND LHL AND SOLVE IT U WILL FAOUND THAT RHL!=LHL

SO ANSWER WILL BE (C)
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