Answer:
Hello, please consider the following.
Evaluate the limit of the numerator and the limit of the denominator.
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0
0
Since
0
0
is of indeterminate form, apply L'Hospital's Rule. L'Hospital's Rule states that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives.
lim
x
→
0
e
x
2
−
1
cos
(
x
)
−
1
=
lim
x
→
0
d
d
x
[
e
x
2
−
1
]
d
d
x
[
cos
(
x
)
−
1
]
Find the derivative of the numerator and denominator.
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lim
x
→
0
2
e
x
2
x
−
sin
(
x
)
Evaluate the limit of the numerator and the limit of the denominator.
Tap for more steps...
0
0
Since
0
0
is of indeterminate form, apply L'Hospital's Rule. L'Hospital's Rule states that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives.
lim
x
→
0
2
e
x
2
x
−
sin
(
x
)
=
lim
x
→
0
d
d
x
[
2
e
x
2
x
]
d
d
x
[
−
sin
(
x
)
]
Find the derivative of the numerator and denominator.
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lim
x
→
0
4
e
x
2
x
2
+
2
e
x
2
−
cos
(
x
)
Take the limit of each term.
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4
e
(
lim
x
→
0
x
)
2
⋅
(
lim
x
→
0
x
)
2
+
2
e
(
lim
x
→
0
x
)
2
−
cos
(
lim
x
→
0
x
)
Evaluate the limits by plugging in
0
for all occurrences of
x
.
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4
e
0
2
⋅
0
2
+
2
e
0
2
−
cos
(
0
)
Simplify the answer.
Thank you
Explanation: