JEE MATHEMATICS MCQ: LIMITS

1 Easy

1 of 1

Application +4 | -1

Let

x_n=\frac{1+2+3+\cdots+n}{n^2}

for

n=1,2,3, \ldots

. Then the value of

\lim _{n \rightarrow \infty} x_n

View answer and explanation

Correct option: A

\frac{1}{2}

Option A (

\frac{1}{2}

) is correct because the sum of the first n natural numbers is

\frac{n(n+1)}{2}

, and dividing by

n^2

and taking the limit as

n

approaches infinity yields

\frac{1}{2}

More Questions from LIMITS

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\lim\limits_{h \to 0} 2\left\{ {\frac{{\sqrt 3 \sin \left( {\frac{\pi }{6} + h} \right) - \cos \left( {\frac{\pi }{6} + h} \right)}}{{\sqrt 3 h\left( {\sqrt 3 \cos \,h - \sin \,h} \right)}}} \right\}

\lim \limits_{{x \rightarrow \infty}} x\left(e^{1 / x}-1\right)

f(x)=\frac{(x+1)^7 \sqrt{1+x^2}}{\left(x^2-x+1\right)^6}

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\lim\limits_{n \to \infty } {\left( {1 + \sin \frac{a}{n}} \right)^n}