JEE MAIN - Mathematics (2023 - 25th January Morning Shift - No. 5)
The value of $$\mathop {\lim }\limits_{n \to \infty } {{1 + 2 - 3 + 4 + 5 - 6\, + \,.....\, + \,(3n - 2) + (3n - 1) - 3n} \over {\sqrt {2{n^4} + 4n + 3} - \sqrt {{n^4} + 5n + 4} }}$$ is :
$${3 \over {2\sqrt 2 }}$$
$${3 \over 2}(\sqrt 2 + 1)$$
$$3(\sqrt 2 + 1)$$
$${{\sqrt 2 + 1} \over 2}$$
Explanation
$$
\begin{aligned}
& I=\lim _{n \rightarrow \infty} \frac{(1+2+3+\ldots+3 n)-2(3+6+9+. .+3 n)}{\sqrt{2 n^4+4 n+3}-\sqrt{n^4+5 n+4}} \\\\
& =\lim _{n \rightarrow \infty} \frac{\frac{3 n(3 n+1)}{2}-6 \frac{n(n+1)}{2}}{\left(\sqrt{2 n^4+4 n+3}-\sqrt{n^4+5 n+4}\right)} \\\\
& =\lim _{n \rightarrow \infty} \frac{3 n(n-1)\left[\sqrt{2 n^4+4 n+3}+\sqrt{n^4+5 n+4}\right]}{2 \cdot\left[\left(2 n^4+4 n-3\right)-\left(n^4+5 n+4\right)\right]} \\\\
& =\lim _{n \rightarrow \infty} \frac{3 \cdot 1 \cdot\left(1-\frac{1}{n}\right)\left[\sqrt{2+\frac{4}{n^3}+\frac{3}{n^4}}+\sqrt{1+\frac{5}{n^3}+\frac{4}{n^4}}\right]}{2\left[1-\frac{1}{n^3}-\frac{7}{n^4}\right]} \\\\
& =\frac{3(\sqrt{2}+1)}{2} \\
&
\end{aligned}
$$
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