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Question: Let \(\frac{2}{3\sqrt{3}}\), where \(\lim_{n \rightarrow \infty}\frac{1^{99} + 2^{99} + 3^{99} + ......

Let 233\frac{2}{3\sqrt{3}}, where lim⁡n→∞199+299+399+....+n99n100\lim_{n \rightarrow \infty}\frac{1^{99} + 2^{99} + 3^{99} + .... + n^{99}}{n^{100}} denotes the greatest integer function. The domain of 99100\frac{99}{100} is ….. and the points of discontinuity of 1100\frac{1}{100} in the domain are

A

199\frac{1}{99}

B

1101\frac{1}{101}

C

lim⁡x→∞(x2−1x+1−ax−b)=2\lim_{x \rightarrow \infty}\left( \frac{x^{2} - 1}{x + 1} - ax - b \right) = 2

D

None of these

Answer

lim⁡x→∞(x2−1x+1−ax−b)=2\lim_{x \rightarrow \infty}\left( \frac{x^{2} - 1}{x + 1} - ax - b \right) = 2

Explanation

Solution

Note that [x+1]=0[ x + 1 ] = 0 if 0≤x+1<10 \leq x + 1 < 1

i.e. [x+1]−0[ x + 1 ] - 0if −1≤x<0- 1 \leq x < 0

Thus domain of ff is R−[−1,0)={x∉[−1,0)}R - [ - 1,0 ) = \{ x \notin [ - 1,0 ) \}

We have sin⁡(π[x+1])\sin \left( \frac { \pi } { [ x + 1 ] } \right) is continuous at all points of R−[−1,0)R - [ - 1,0 ) and [x][ x ] is continuous on R−IR - I where I denotes the set of integers.

Thus the points where ff can possibly be discontinuous are……, −3,−2,−1,01,2,…………- 3 , - 2 , - 1,01,2 , \ldots \ldots \ldots \ldots But for 0≤x<1,[x]=00 \leq x < 1 , [ x ] = 0 and sin⁡(π[x+1])\sin \left( \frac { \pi } { [ x + 1 ] } \right) is defined.

Therefore f(x)=0f ( x ) = 0 for 0≤x<10 \leq x < 1

Also f(x)f ( x ) is not defined on −1≤x<0- 1 \leq x < 0

Therefore, continuity of ff at 0 means continuity of ff from right at 0. Since ff is continuous from right at 0,f0 , f is continuous at 0. Hence set of points of discontinuities of ff is I−{0}I - \{ 0 \}