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Question: If \(\frac{1}{3}\) for \(x\) \(\lim_{n \rightarrow \infty}\left\lbrack \frac{1}{1 - n^{2}} + \frac{...

If 13\frac{1}{3} for xx lim⁡n→∞[11−n2+21−n2+......+n1−n2]\lim_{n \rightarrow \infty}\left\lbrack \frac{1}{1 - n^{2}} + \frac{2}{1 - n^{2}} + ...... + \frac{n}{1 - n^{2}} \right\rbrack5 and f is continuous at −12- \frac{1}{2} then 12\frac{1}{2}

A

0

B

5

C

10

D

25

Answer

0

Explanation

Solution

f(5)=lim⁡x→5f(x)f ( 5 ) = \lim _ { x \rightarrow 5 } f ( x ) =lim⁡x→5x2−10x+25x2−7x+10= \lim _ { x \rightarrow 5 } \frac { x ^ { 2 } - 10 x + 25 } { x ^ { 2 } - 7 x + 10 }

=lim⁡x→5(x−5)2(x−2)(x−5)=5−55−2=0= \lim _ { x \rightarrow 5 } \frac { ( x - 5 ) ^ { 2 } } { ( x - 2 ) ( x - 5 ) } = \frac { 5 - 5 } { 5 - 2 } = 0.