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Joint Entrance Examination
Mathematics
Limits, Continuity and Differentiability
Question 375
Question 375 of 401
Medium
MCQ
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 :
A
\[{3 \over {2\sqrt 2 }}\]
B
\[{3 \over 2}(\sqrt 2 + 1)\]
C
\[3(\sqrt 2 + 1)\]
D
\[{{\sqrt 2 + 1} \over 2}\]
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Official solution
\[
I
=
lim
n
→
∞
(
1
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2
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3
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…
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3
n
)
−
2
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n
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n
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n
4
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5
n
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4
=
lim
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∞
3
n
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3
n
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2
−
6
n
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n
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1
)
2
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n
4
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4
n
+
3
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n
4
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5
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4
)
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lim
n
→
∞
3
n
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n
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)
[
2
n
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]
2
⋅
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−
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lim
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∞
3
⋅
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⋅
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n
)
[
2
+
4
n
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+
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]
2
[
1
−
1
n
3
−
7
n
4
]
=
3
(
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+
1
)
2
\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}
I
=
n
→
∞
lim
2
n
4
+
4
n
+
3
−
n
4
+
5
n
+
4
(
1
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3
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−
2
(
3
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+
..
+
3
n
)
=
n
→
∞
lim
(
2
n
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4
n
+
3
−
n
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n
+
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3
n
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−
6
2
n
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=
n
→
∞
lim
2
⋅
[
(
2
n
4
+
4
n
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3
)
−
(
n
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+
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]
3
n
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n
−
1
)
[
2
n
4
+
4
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+
3
+
n
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+
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n
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]
=
n
→
∞
lim
2
[
1
−
n
3
1
−
n
4
7
]
3
⋅
1
⋅
(
1
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1
)
[
2
+
n
3
4
+
n
4
3
+
1
+
n
3
5
+
n
4
4
]
=
2
3
(
2
+
1
)
\]
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