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Joint Entrance Examination
Mathematics
Limits, Continuity and Differentiability
Question 112
Question 112 of 401
Medium
MCQ
lim
x
→
∞
(
2
x
2
−
3
x
+
5
3
x
2
+
5
x
+
4
)
x
/
2
\displaystyle\lim_{x\to\infty}\left(\frac{2x^2-3x+5}{3x^2+5x+4}\right)^{x/2}
x
→
∞
lim
(
3
x
2
+
5
x
+
4
2
x
2
−
3
x
+
5
)
x
/2
is equal to:
A
2
e
3
/
2
2e^{3/2}
2
e
3/2
B
2
3
e
\dfrac{2}{3\sqrt e}
3
e
2
C
2
e
3
/
2
2e\sqrt{3/2}
2
e
3/2
D
3
e
\sqrt{3}e
3
e
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Official solution
Write
2
x
2
−
3
x
+
5
3
x
2
+
5
x
+
4
=
2
−
3
x
+
5
x
2
3
+
5
x
+
4
x
2
\dfrac{2x^2-3x+5}{3x^2+5x+4}=\dfrac{2-\frac3x+\frac5{x^2}}{3+\frac5x+\frac4{x^2}}
3
x
2
+
5
x
+
4
2
x
2
−
3
x
+
5
=
3
+
x
5
+
x
2
4
2
−
x
3
+
x
2
5
, and use
(
1
−
a
x
)
x
/
2
→
e
−
a
/
2
\left(1-\dfrac{a}{x}\right)^{x/2}\to e^{-a/2}
(
1
−
x
a
)
x
/2
→
e
−
a
/2
for the numerator/denominator factors near
1
1
1
: \[\lim_{x\to\infty}\frac23\cdot\frac{e^{-1/6}}{e^{1/3}}=\frac23\,e^{-1/2}=\frac{2}{3\sqrt e}\]
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