n → ∞ lim ( 3 8 n + 4 n + 2 n − 2 n ) = ?
Give your answer to 3 decimal places.
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Same approach :-)
Nice solution. i used L'Hospital's rule!
i use the rationalize method to do this
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Could you post your solution too? ;)
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don't know how to post here :(
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@Wai Lam William Cheung
–
Credit to
@Wai Lam William Cheung
:
x
→
∞
lim
(
3
8
n
+
4
n
+
2
n
−
2
n
)
=
x
→
∞
lim
1
3
8
n
+
4
n
+
2
n
−
2
n
(
8
n
+
4
n
+
2
n
)
2
/
3
+
(
8
n
+
4
n
+
2
n
)
1
/
3
(
2
n
)
+
(
2
n
)
2
(
8
n
+
4
n
+
2
n
)
2
/
3
+
(
8
n
+
4
n
+
2
n
)
1
/
3
(
2
n
)
+
(
2
n
)
2
=
x
→
∞
lim
(
8
n
+
4
n
+
2
n
)
2
/
3
+
(
8
n
+
4
n
+
2
n
)
1
/
3
(
2
n
)
+
(
2
n
)
2
(
8
n
+
4
n
+
2
n
)
−
8
n
=
x
→
∞
lim
(
8
n
+
4
n
+
2
n
)
2
/
3
+
(
8
n
+
4
n
+
2
n
)
1
/
3
(
2
n
)
+
(
2
n
)
2
4
n
+
2
n
=
x
→
∞
lim
(
1
+
(
2
1
)
n
+
(
4
1
)
n
)
2
/
3
+
(
1
+
(
2
1
)
n
+
(
4
1
)
n
)
1
/
3
+
1
1
+
(
2
1
)
n
=
1
+
1
+
1
1
=
3
1
Let x = ( 8 n + 4 n + 2 n ) 3 1 and y = 2 n = ( 8 n ) 3 1 . Since x 3 − y 3 = ( x + y ) ( x 2 + x y + y 2 ) , we have x − y = x 2 + x y + y 2 x 3 − y 3 , which means that ( 8 n + 4 n + 2 n ) 3 1 − 2 n = ( 8 n + 4 n + 2 n ) 3 2 + ( 8 n + 4 n + 2 n ) 3 1 × 2 n + 4 n 4 n + 2 n = ( 1 + ( 2 1 ) n + ( 4 1 ) n ) 3 2 + 1 + 1 ( 1 + ( 2 1 ) n + ( 4 1 ) n ) 3 1 1 + ( 2 1 ) n
This means that n → ∞ lim ( ( 8 n + 4 n + 2 n ) 3 1 − 2 n ) = n → ∞ lim ( 1 + ( 2 1 ) n + ( 4 1 ) n ) 3 2 + 1 + 1 ( 1 + ( 2 1 ) n + ( 4 1 ) n ) 3 1 1 + ( 2 1 ) n = 3 1
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Let the limit be H . Substitute ⋆ 2 n = x followed by using binomial approximation ⋆ ( 1 + α ) n = 1 + n α ( as α → 0 ) .
H = = = = x → ∞ lim ( 3 x 3 + x 2 + x − x ) x → ∞ lim ( x ( 3 1 + x 1 + x 2 1 − 1 ) ) x → ∞ lim ( x ( 1 + 3 1 ( x 1 + x 2 1 ) − 1 ) ) x → ∞ lim ⎝ ⎜ ⎛ 3 1 + 0 3 x 1 ⎠ ⎟ ⎞ ≈ 0 . 3 3 3