Definite Integration Question 6
Question 6 - 25 January - Shift 1
The value of
$ Lim _{n \to \infty} \frac{1+2-3+4+5-6+\ldots+(3 n-2)+(3 n-1)-3 n}{\sqrt{2 n^{4}+4 n+3}-\sqrt{n^{4}+5 n+4}} $
is :
(1) $\frac{\sqrt{2}+1}{2}$
(2) $3(\sqrt{2}+1)$
(3) $\frac{3}{2}(\sqrt{2}+1)$
(4) $\frac{3}{2 \sqrt{2}}$
Show Answer
Answer: (3)
Solution:
Formula: Sum of first n natural numbers, Algebra of limits (5)
$\begin{aligned} & \lim {n \rightarrow \infty} \frac{\sum{r=1}^n ((3 r-1)+(3 r-2)-3 r)}{\sqrt{2 n^4+4 n+3}-\sqrt{n^4+5 n+4}} \\ &=\lim {n \rightarrow \infty} \frac{\sum{r=1}^n(3 r-3)}{\sqrt{2 n^4+4 n+3}-\sqrt{n^4+5 n+4}}\end{aligned}$
$\begin{aligned} & =\lim _{n \rightarrow \infty} \frac{3 \cdot n \frac{(n+1)}{2}-3}{n^2 \sqrt{2+\frac{4}{n^3}+\frac{3}{n^4}}-\sqrt{1+\frac{5}{n^3}+\frac{4}{n^4}}} \\ & =\lim _{n \rightarrow \infty} \frac{\frac{3}{2}\left(1+\frac{1}{n}\right)-\frac{3}{n^2}}{\sqrt{2+\frac{4}{n^3}+\frac{3}{n^4}}-\sqrt{1+\frac{5}{n^3}+\frac{4}{n^4}}} \\ & \end{aligned}$
$\begin{aligned} & =\frac{\frac{3}{2}(1+0)-0}{\sqrt{2+0+0}-\sqrt{1+0+0}} \\ & =\frac{\frac{3}{2}}{\sqrt{2}-1}\end{aligned}$
$\begin{aligned} & =\frac{\frac{3}{2}(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)} \\ & =\frac{\frac{3}{2}(\sqrt{2}+1)}{2-1}=\frac{3}{2}(\sqrt{2}+1)\end{aligned}$
