mathop {lim }limits_{x to 0} x^2(1+2+3+...+[f | vedclass

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Question

(B) Let $t = \frac{1}{|x|}$. As $x 	o 0$,$t 	o \infty$.The expression becomes $\mathop {\lim }\limits_{t 	o \infty } \frac{1}{t^2} (1 + 2 + 3 + ...

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$\frac{1}{2}$

Vedclass · Answer

(B) Let $t = \frac{1}{|x|}$. As $x 	o 0$,$t 	o \infty$.The expression becomes $\mathop {\lim }\limits_{t 	o \infty } \frac{1}{t^2} (1 + 2 + 3 + ... + [t])$.Using the sum formula for the first $n$ natural numbers,$1+2+...+n = \frac{n(n+1)}{2}$,we have:$\mathop {\lim }\limits_{t 	o \infty } \frac{[t]([t]+1)}{2t^2}$.Since $\mathop {\lim }\limits_{t 	o \infty } \frac{[t]}{t} = 1$,the limit is $\mathop {\lim }\limits_{t 	o \infty } \frac{1}{2} \cdot \frac{[t]}{t} \cdot \frac{[t]+1}{t} = \frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2}$.

$\mathop {\lim }\limits_{x \to 0} x^2(1+2+3+...+[\frac{1}{|x|}])$ is equal to (where $[.]$ denotes the greatest integer function).

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