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If f(x) =   f (x)= \left\{\begin{matrix} \frac{\sin \left \{ \cos x \right \}}{x-\pi/2} & , x \neq \pi/2 \\ 1& , x = \pi/2 \end{matrix}\right.,  where {.}  represents the fractional part function,

then  

                   

Option: 1

f(x) is continuous at   x= \pi /2 


Option: 2

\lim_{x\rightarrow \pi/2 }  f(x)  exists , but f is not continuous at x= \pi /2


Option: 3

\lim_{x\rightarrow \pi/2 }  f(x)  does not exists


Option: 4

\lim_{x\rightarrow \pi/2 }  f(x) =1 


Answers (1)

best_answer

 

Continuity at a point -

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

1.  f(a) is defined  : at  x = a.

2. \lim_{x\rightarrow a}\:f(x)\:exists\:means\:limit\:x\rightarrow a

of  f(x) at  x = a exists from left and right.

3. \lim_{x\rightarrow a}\:f(x)=f(a)\:then\:the\:limit\:equals \:the\:value\:at\:x=a

-

 

 

Continuity from Left -

The function f(x) is said to be continuous from left at x = a if 

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

-

 

 

Continuity from Right -

The function f(x) is said to be continuous from right at 

x=a:if\:\lim_{x\rightarrow a^{+}}\:f(x)=f(a)

-

 

 

R.H.L = \lim_{h\rightarrow 0 ^+ } f ( \pi /2+h ) = \lim_{h\rightarrow 0 ^+ } \frac{\sin (1- \sin h )}{h }\rightarrow \infty \\\\L.H.L = \lim_{h\rightarrow 0 ^- } f ( \pi /2-h ) = \lim_{h\rightarrow 0 ^- } \frac{\sin (\sin h )}{-h }\\\\ \lim_{h \rightarrow 0 ^-} \left ( \frac{\sin ( \sin h )}{\sin h }\times \frac{\sin h }{-h} \right ) = 1 * -1 = -1

L.H.L \neq  R.H.L 

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Nehul

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