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NDA 2020
Exam
NDA
Year
2020
Subject
Mathematics
Chapter
Limits
NDA 2020 · Mathematics · Limits
What is _ x arrow 0 ( x (1-x))/(x^2) equal to?
What is
lim
x
→
0
sin
x
log
(
1
−
x
)
x
2
\lim _{x \rightarrow 0} \frac{\sin x \log (1-x)}{x^2}
lim
x
→
0
x
2
s
i
n
x
l
o
g
(
1
−
x
)
equal to?
Options
A
-1
B
Zero
C
−
e
-e
−
e
D
−
1
e
-\frac{1}{e}
−
e
1
Correct answer
A. -1
Step-by-step solution
lim
x
→
0
sin
x
⋅
log
(
1
−
x
)
x
2
\lim _{x \rightarrow 0} \frac{\sin x \cdot \log (1-x)}{x^2}
lim
x
→
0
x
2
s
i
n
x
⋅
l
o
g
(
1
−
x
)
This is in
0
0
\frac{0}{0}
0
0
form, by using L' Hospital rule.
lim
x
→
0
sin
x
⋅
log
(
1
−
x
)
x
2
=
lim
x
→
0
cos
x
⋅
log
(
1
−
x
)
+
sin
x
⋅
1
(
1
−
x
)
(
−
1
)
2
x
\begin{aligned} & \lim _{x \rightarrow 0} \frac{\sin x \cdot \log (1-x)}{x^2} \\ & =\lim _{x \rightarrow 0} \frac{\cos x \cdot \log (1-x)+\sin x \cdot \frac{1}{(1-x)}(-1)}{2 x}\end{aligned}
x
→
0
lim
x
2
sin
x
⋅
lo
g
(
1
−
x
)
=
x
→
0
lim
2
x
cos
x
⋅
lo
g
(
1
−
x
)
+
sin
x
⋅
(
1
−
x
)
1
(
−
1
)
=
lim
x
→
0
log
(
1
−
x
)
⋅
(
−
sin
x
)
−
cos
x
(
1
−
x
)
−
1
(
1
−
x
)
⋅
cos
x
−
sin
x
(
1
−
x
)
2
2
=
0
−
1
−
1
−
0
2
=
−
2
2
=
−
1
\begin{aligned} & =\lim _{x \rightarrow 0} \frac{\log (1-x) \cdot(-\sin x)-\frac{\cos x}{(1-x)}-\frac{1}{(1-x)} \cdot \cos x-\frac{\sin x}{(1-x)^2}}{2} \\ & =\frac{0-1-1-0}{2}=\frac{-2}{2}=-1\end{aligned}
=
x
→
0
lim
2
lo
g
(
1
−
x
)
⋅
(
−
sin
x
)
−
(
1
−
x
)
c
o
s
x
−
(
1
−
x
)
1
⋅
cos
x
−
(
1
−
x
)
2
s
i
n
x
=
2
0
−
1
−
1
−
0
=
2
−
2
=
−
1
$
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