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Question: The left hand derivative of \(f ( x ) = [ x ]\) \(\sqrt{2}\) at \(\frac{1}{\sqrt{2}}\) (k is an inte...

The left hand derivative of f(x)=[x]f ( x ) = [ x ] 2\sqrt{2} at 12\frac{1}{\sqrt{2}} (k is an integer), is

A

lim⁡x→1\lim_{x \rightarrow 1}

B

sin⁡∣∣∣x∣−2∣−3∣\sin\left| ||x| - 2| - 3 \right|

C

lim⁡x→0\lim_{x \rightarrow 0}

D

lim⁡n→∞\lim_{n \rightarrow \infty}

Answer

lim⁡x→1\lim_{x \rightarrow 1}

Explanation

Solution

f(x)=[x]f ( x ) = [ x ] sin⁡(πx)\sin ( \pi x )

If x is just less than k, [x] = k – 1. ∴ , when x<kx < k ∀k∈I\forall k \in I

Now L.H.D. at x=kx = k,

=lim⁡x→k(k−1)sin⁡(πx)−ksin⁡(πk)x−k\lim _ { x \rightarrow k } \frac { ( k - 1 ) \sin ( \pi x ) - k \sin ( \pi k ) } { x - k } = lim⁡x→k(k−1)sin⁡(πx)(x−k)\lim _ { x \rightarrow k } \frac { ( k - 1 ) \sin ( \pi x ) } { ( x - k ) }

[as sin⁡(πk)=0,k∈\sin ( \pi k ) = 0 , k \in integer]

= lim⁡h→0(k−1)sin⁡(π(k−h))−h\lim _ { h \rightarrow 0 } \frac { ( k - 1 ) \sin ( \pi ( k - h ) ) } { - h } [Let x=(k−h)x = ( k - h ) ]

= lim⁡h→0(k−1)(−1)k−1sin⁡hπ−h\lim _ { h \rightarrow 0 } \frac { ( k - 1 ) ( - 1 ) ^ { k - 1 } \sin h \pi } { - h }

= lim⁡h→0(k−1)(−1)k−1sin⁡hπhπ×(−π)\lim _ { h \rightarrow 0 } ( k - 1 ) ( - 1 ) ^ { k - 1 } \frac { \sin h \pi } { h \pi } \times ( - \pi ) = (k−1)(−1)kπ( k - 1 ) ( - 1 ) ^ { k } \pi

= (−1)k(k−1)π( - 1 ) ^ { k } ( k - 1 ) \pi.