The value of operatorname{Lim}_{n rightarrow | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The numerator is a sum of $n$ groups of the form $(3k-2) + (3k-1) - 3k = 3k-3$ for $k=1$ to $n$.Sum $= \sum_{k=1}^{n} (3k-3) = 3 \sum_{k=1}^{n} (k

Vedclass · Accepted Answer

$\frac{3}{2}(\sqrt{2}+1)$

Vedclass · Answer

(C) The numerator is a sum of $n$ groups of the form $(3k-2) + (3k-1) - 3k = 3k-3$ for $k=1$ to $n$.Sum $= \sum_{k=1}^{n} (3k-3) = 3 \sum_{k=1}^{n} (k-1) = 3 \frac{(n-1)n}{2} = \frac{3n^2-3n}{2}$.Now,consider the limit: $\operatorname{Lim}_{n ightarrow \infty} \frac{\frac{3n^2-3n}{2}}{\sqrt{2n^4+4n+3}-\sqrt{n^4+5n+4}}$.Divide numerator and denominator by $n^2$: $\operatorname{Lim}_{n}$ ${ightarrow \infty} \frac{\frac{3}{2} - \frac{3}{2n}}{\sqrt{2+\frac{4}{n^3}+\frac{3}{n^4}}-\sqrt{1+\frac{5}{n^3}+\frac{4}{n^4}}}$.As $n ightarrow \infty$,the expression approaches $\frac{3/2}{\sqrt{2}-1}$.Rationalizing the denominator: $\frac{3}{2(\sqrt{2}-1)} 	imes \frac{\sqrt{2}+1}{\sqrt{2}+1} = \frac{3(\sqrt{2}+1)}{2(2-1)} = \frac{3}{2}(\sqrt{2}+1)$.

The value of $\operatorname{Lim}_{n \rightarrow \infty} \frac{1+2-3+4+5-6+\ldots+(3n-2)+(3n-1)-3n}{\sqrt{2n^4+4n+3}-\sqrt{n^4+5n+4}}$ is:

Similar Questions

If $f(x)$ satisfies $97 f(x) + m f\left(\frac{1}{x}\right) = 0$,where $f(x) = \lim_{n \rightarrow \infty} n(x^{1/n} - 1)$ for $x > 0$,then the value of $m$ is:

Evaluate the given limit: $\mathop {\lim }\limits_{x \to 2} \frac{3x^{2}-x-10}{x^{2}-4}$

$\lim _{x \rightarrow 0} \frac{\sqrt{1+x \sin x}-\sqrt{\cos x}}{\tan ^2 2 x}=$

$\lim _{x \rightarrow \infty}\left[\frac{x^2+x+3}{x^2-x+2}\right]^x$ is equal to

$\lim _{x \rightarrow 0} \frac{(\operatorname{cosec} x-\cot x)(e^x-e^{-x})}{\sqrt{3}-\sqrt{2+\cos x}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module