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Byju's Answer
Standard XII
Mathematics
Factorization Method Form to Remove Indeterminate Form
Evaluate: l...
Question
Evaluate:
lim
x
→
1
√
1
−
cos
2
(
x
−
1
)
x
−
1
A
exists and it equals
√
2
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B
exists and it equals
−
√
2
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C
does not exist because
x
−
1
→
0
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D
does not exist because left hand limit is not equal to right hand limit
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Solution
The correct option is
D
exists and it equals
√
2
lim
x
→
1
√
1
−
cos
2
(
x
−
1
)
x
−
1
(
÷
f
o
r
m
)
=
lim
x
→
1
√
2
sin
2
(
x
−
1
)
x
−
1
..........
(
∵
1
−
c
o
s
2
x
=
2
s
i
n
2
x
)
=
lim
x
→
1
√
2
sin
(
x
−
1
)
x
−
1
=
√
2
[
lim
x
→
0
sin
x
x
=
1
]
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0
Similar questions
Q.
Evaluate the left hand and right hand limits of the function defined by
f
(
x
)
=
{
1
+
x
2
,
if
0
≤
x
≤
1
2
−
x
,
if
x
>
1
at
x
=
1
. Also show that
lim
x
→
1
f
(
x
)
does not exist.
Q.
Let
lim
x
→
a
f
(
x
)
exists but it is not equal to f (a). Then f(x) is discontinuous at x
=
a and a is called a removable discontinuity. If
lim
x
→
a
−
f
(
x
)
=
l
a
n
d
lim
x
→
a
+
f
(
x
)
=
m
exist but
l
≠
m
.
Then a is called a jump discontinuity. If one of the limits (left hand limit or right hand limit ) does not exist, then a is called an infinite discontinuity.
Let f(x)
⎧
⎪ ⎪ ⎪ ⎪
⎨
⎪ ⎪ ⎪ ⎪
⎩
2
|
x
|
,
x
≤
−
1
2
x
,
−
1
≤
x
≤
0
x
+
1
,
0
<
x
≤
1
2
x
>
1
Then
f
(
x
)
at
Q.
Assertion :
lim
x
→
0
e
1
/
x
−
1
e
1
/
x
+
1
does not exist. Reason:
lim
x
→
0
+
e
1
/
x
−
1
e
1
/
x
+
1
does not exist.
Q.
Assertion :
lim
x
→
0
√
1
−
cos
2
x
x
does not exist. Reason:
|
sin
x
|
=
⎧
⎪
⎨
⎪
⎩
sin
x
;
0
<
x
<
π
2
−
sin
x
;
−
π
2
<
x
<
0
Q.
Assertion :If
lim
x
→
0
{
f
(
x
)
+
sin
x
x
}
does not exist, then
lim
x
→
0
f
(
x
)
does not exist. Reason:
lim
x
→
0
sin
x
x
exists and has value 1.
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