Limits 3.0

Calculus Level 4

The value of limx0x2(1+2+3+......[1x])=A \displaystyle \lim_{x \to 0} |x|^{2}(1+2+3 +......[\frac{1}{|x|}])=A . where |x| means absolute value of x and [.] means Greatest Integer function.Find 1A \frac{1}{A}

×

Problem Loading...

Note Loading...

Set Loading...