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Joint Entrance Examination
Mathematics
Limits, Continuity and Differentiability
Question 2
Question 2 of 401
Medium
MCQ
l
i
m
x
→
0
(
x
2
sin
2
x
x
2
−
sin
2
x
)
\underset{x\to 0}{lim}(\frac{{x}^{2}{\sin }^{2}x}{{x}^{2}-{\sin }^{2}x})
x
→
0
l
im
(
x
2
−
s
i
n
2
x
x
2
s
i
n
2
x
)
A
2
B
3
C
4
D
6
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Solutions · 1
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Official solution
l
i
m
x
→
0
x
2
sin
2
x
x
2
−
sin
2
x
\underset{x\to 0}{lim}\frac{{x}^{2}{\sin }^{2}x}{{x}^{2}-{\sin }^{2}x}
x
→
0
l
im
x
2
−
s
i
n
2
x
x
2
s
i
n
2
x
Applying expansion
l
i
m
x
→
0
x
2
sin
2
x
x
2
−
(
x
−
x
3
3
!
+
…
)
2
l
i
m
x
→
0
x
2
sin
2
x
x
2
−
x
2
+
2
x
4
3
!
…
=
3
!
2
=
3
\begin{gathered} \underset{x\to 0}{lim}\frac{{x}^{2}{\sin }^{2}x}{{x}^{2}-{(x-\frac{{x}^{3}}{3!}+\ldots )}^{2}} \\ \underset{x\to 0}{lim}\frac{{x}^{2}{\sin }^{2}x}{{x}^{2}-{x}^{2}+\frac{2{x}^{4}}{3!}\ldots }=\frac{3!}{2}=3 \end{gathered}
x
→
0
l
im
x
2
−
(
x
−
3
!
x
3
+
…
)
2
x
2
sin
2
x
x
→
0
l
im
x
2
−
x
2
+
3
!
2
x
4
…
x
2
sin
2
x
=
2
3
!
=
3
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