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Exams
Joint Entrance Examination
Mathematics
Indefinite Integrals
Question 363
Question 363 of 416
Medium
MCQ
lim
n
→
∞
1
4
+
2
4
+
3
4
+
⋯
+
n
4
n
5
−
lim
n
→
∞
1
3
+
2
3
+
3
3
+
⋯
+
n
3
n
5
\displaystyle\lim_{n\to\infty} \dfrac{1^4+2^4+3^4+\cdots+n^4}{n^5} - \lim_{n\to\infty} \dfrac{1^3+2^3+3^3+\cdots+n^3}{n^5}
n
→
∞
lim
n
5
1
4
+
2
4
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3
4
+
⋯
+
n
4
−
n
→
∞
lim
n
5
1
3
+
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3
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3
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⋯
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n
3
A
1
5
\dfrac15
5
1
B
1
30
\dfrac{1}{30}
30
1
C
Zero
D
1
4
\dfrac14
4
1
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Official solution
=
lim
n
→
∞
1
n
∑
r
=
1
n
(
r
n
)
4
−
lim
n
→
∞
1
n
⋅
lim
n
→
∞
1
n
∑
r
=
1
n
(
r
n
)
3
= \lim_{n\to\infty}\dfrac1n\sum_{r=1}^n\left(\dfrac{r}{n}\right)^4 - \lim_{n\to\infty}\dfrac1n\cdot\lim_{n\to\infty}\dfrac1n\sum_{r=1}^n\left(\dfrac{r}{n}\right)^3
=
lim
n
→
∞
n
1
∑
r
=
1
n
(
n
r
)
4
−
lim
n
→
∞
n
1
⋅
lim
n
→
∞
n
1
∑
r
=
1
n
(
n
r
)
3
=
∫
0
1
x
4
d
x
−
0
×
∫
0
1
x
3
d
x
=
[
x
5
5
]
0
1
−
0
=
1
5
= \int_0^1 x^4\,dx - 0\times\int_0^1 x^3\,dx = \left[\dfrac{x^5}{5}\right]_0^1 - 0 = \dfrac15
=
∫
0
1
x
4
d
x
−
0
×
∫
0
1
x
3
d
x
=
[
5
x
5
]
0
1
−
0
=
5
1
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