The value of mathop {lim }limits_{n to infty | vedclass

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(B) We know that the sum of the first $n$ natural numbers is given by $\frac{n(n+1)}{2}$.Substituting this into the limit,we get:$\mathop {\lim }\limi

Vedclass · Accepted Answer

$\frac{1}{2}$

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(B) We know that the sum of the first $n$ natural numbers is given by $\frac{n(n+1)}{2}$.Substituting this into the limit,we get:$\mathop {\lim }\limits_{n 	o \infty } \frac{n(n + 1)}{2({n^2} + 100)}$$= \mathop {\lim }\limits_{n 	o \infty } \frac{{n^2} + n}{2{n^2} + 200}$Dividing the numerator and the denominator by ${n^2}$:$= \mathop {\lim }\limits_{n 	o \infty } \frac{1 + \frac{1}{n}}{2 + \frac{200}{n^2}}$As $n 	o \infty$,$\frac{1}{n} 	o 0$ and $\frac{200}{n^2} 	o 0$.Therefore,the limit is $\frac{1 + 0}{2 + 0} = \frac{1}{2}$.

The value of $\mathop {\lim }\limits_{n \to \infty } \frac{{1 + 2 + 3 + .... + n}}{{{n^2} + 100}}$ is equal to:

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