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Joint Entrance Examination
Mathematics
Limits, Continuity and Differentiability
Question 34
Question 34 of 401
Medium
MCQ
l
i
m
x
→
0
e
−
(
1
+
2
x
)
1
2
x
x
\underset{x\to 0}{lim}\frac{e-(1+2x{)}^{\frac{1}{2x}}}{x}
x
→
0
l
im
x
e
−
(
1
+
2
x
)
2
x
1
is equal to
A
−
2
e
\frac{-2}{e}
e
−
2
B
e
−
e
2
e-{e}^{2}
e
−
e
2
C
0
D
e
e
e
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Official solution
l
i
m
x
→
0
e
−
(
1
+
2
x
)
1
2
x
x
\underset{x\to 0}{lim}\frac{e-(1+2x{)}^{\frac{1}{2x}}}{x}
x
→
0
l
im
x
e
−
(
1
+
2
x
)
2
x
1
Using expansion
=
l
i
m
x
→
0
e
−
e
[
1
−
2
x
2
+
11
×
4
x
2
24
+
…
]
x
=
l
i
m
x
→
0
(
e
−
11
x
6
e
+
…
)
=
e
\begin{gathered} =\underset{x\to 0}{lim}\frac{e-e[1-\frac{2x}{2}+\frac{11\times 4{x}^{2}}{24}+\ldots ]}{x} \\ =\underset{x\to 0}{lim}(e-\frac{11x}{6}e+\ldots )=e \end{gathered}
=
x
→
0
l
im
x
e
−
e
[
1
−
2
2
x
+
24
11
×
4
x
2
+
…
]
=
x
→
0
l
im
(
e
−
6
11
x
e
+
…
)
=
e
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