1-Lim-n-1-n-2-2-n-2-3-n-2-n-1-n-2-2-lim-x-0-3sin5x-x-1-cos4x-x-2-

Question Number 216144 by klipto last updated on 28/Jan/25

1 . {Lim}_{n \to \infty} [\frac{1}{n^{2}} + \frac{2}{n^{2}} + \frac{3}{n^{2}} + \dots + \frac{n + 1}{n^{2}}]

2 . \lim_{x \to 0} {(\frac{3 \sin 5 x}{x})}^{\frac{1 - \cos 4 x}{x^{2}}}

Answered by Ghisom last updated on 28/Jan/25

\lim_{n \to \infty} \underset{k = 1}{\sum^{n + 1}} \frac{k}{n^{2}} = \lim_{n \to \infty} \frac{\underset{k = 1}{\sum^{n + 1}} k}{n^{2}} = \lim_{n \to \infty} \frac{n^{2} + 3 n + 2}{2 n^{2}} = \frac{1}{2}

Answered by mr W last updated on 28/Jan/25

1)

\lim_{n \to \infty} \underset{k = 1}{\sum^{n + 1}} \frac{k}{n^{2}}

= \int_{0}^{1} x d x

= \frac{1}{2}

Commented by klipto last updated on 28/Jan/25

thanks bossman

Answered by mr W last updated on 28/Jan/25

2)

= \lim_{x \to 0} {(\frac{15 \sin 5 x}{5 x})}^{\frac{2 \sin^{2} 2 x}{x^{2}}}

= \lim_{x \to 0} {(\frac{15 \sin 5 x}{5 x})}^{8 {(\frac{\sin 2 x}{2 x})}^{2}}

= {(15 \times 1)}^{8 \times 1^{2}}

= 15^{8}

Commented by klipto last updated on 28/Jan/25

thanks ghisom

Commented by Ghisom last updated on 28/Jan/25

yes . I missed the power \dots

Answered by MrGaster last updated on 29/Jan/25

(1) := \lim_{n \to \infty} [\frac{1 + 2 + 3 + \dots + (n + 1)}{n^{2}}]

= \lim_{n \to \infty} [\frac{(n + 1) (n + 2)}{\frac{2}{n^{2}}}]

= \lim_{n \to \infty} [\frac{n^{2} + 3 n + 2}{2 n^{2}}]

= \frac{1}{2}

(2) := \lim_{x \to 0} {(3 \cdot \frac{\sin 5 x}{5 x} \cdot 5)}^{\frac{1 - (1 - \frac{(4 x^{2})}{2!} + \dots)}{x^{2}}}

= \lim_{x \to 0} {(15 \cdot \frac{\sin 5 x}{5 x})}^{\frac{8 x^{2}}{x^{2}}}

= {\underset{x \to 0}{\lim 15}}^{8} (\lim_{x \to 0} \frac{\sin 5 x}{5 x} = 1)

= 15^{8}

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