what is the value of lim x-0 sin5x/tan6x
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Answered by
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The first thing you should always try with limits is just to enter the x value in the function:
lim
x
→
0
tan
(
6
x
)
sin
(
2
x
)
=
tan
(
6
⋅
0
)
sin
(
2
⋅
0
)
=
tan
(
0
)
sin
(
0
)
=
(
0
0
)
This is an impossible answer, but whenever we find that we have
(
0
0
)
, there's a trick we can use. It's called L'Hôpital's Rule. It states formally that:
lim
x
→
a
f
(
x
)
g
(
x
)
=
lim
x
→
a
f
'
(
x
)
g
'
(
x
)
In this formula,
f
'
(
x
)
means the derivative of f(x) and
g
'
(
x
)
means the derivative of g(x). If you haven't learned about them yet, you can learn them here.
Let's apply this rule to our problem.
f
(
x
)
=
tan
(
6
x
)
f
'
(
x
)
=
d
d
x
tan
(
6
x
)
=
sec
2
(
6
x
)
⋅
d
d
x
(
6
x
)
(chain rule)
=
6
⋅
sec
2
(
6
x
)
=
6
cos
2
(
6
x
)
(since
sec
=
1
cos
)
So that's our first derivative. Now for our second:
g
(
x
)
=
sin
(
2
x
)
g
'
(
x
)
=
d
d
x
sin
(
2
x
)
=
cos
(
2
x
)
⋅
d
d
x
(
2
x
)
(chain rule)
=
2
⋅
cos
(
2
x
)
Now, all we need to do is combine both of them.
lim
x
→
0
tan
(
6
x
)
sin
(
2
x
)
=
lim
x
→
0
(
6
⋅
sec
2
(
6
x
)
2
⋅
cos
(
2
x
)
)
You can bring the number in front of the limit:
=
6
2
lim
x
→
0
(
sec
2
(
6
x
)
cos
(
2
x
)
)
Now we can replace
sec
by
1
cos
=
3
⋅
lim
x
→
0
(
1
cos
2
(
6
x
)
⋅
cos
(
2
x
)
)
Now let's try entering
x
into this formula:
=
3
⋅
(
1
cos
2
(
6
⋅
0
)
⋅
cos
(
2
⋅
0
)
)
=
3
⋅
(
1
cos
2
(
0
)
⋅
cos
(
0
)
)
=
3
⋅
1
(since
cos
(
0
)
=
1
)
=
3
lim
x
→
0
tan
(
6
x
)
sin
(
2
x
)
=
tan
(
6
⋅
0
)
sin
(
2
⋅
0
)
=
tan
(
0
)
sin
(
0
)
=
(
0
0
)
This is an impossible answer, but whenever we find that we have
(
0
0
)
, there's a trick we can use. It's called L'Hôpital's Rule. It states formally that:
lim
x
→
a
f
(
x
)
g
(
x
)
=
lim
x
→
a
f
'
(
x
)
g
'
(
x
)
In this formula,
f
'
(
x
)
means the derivative of f(x) and
g
'
(
x
)
means the derivative of g(x). If you haven't learned about them yet, you can learn them here.
Let's apply this rule to our problem.
f
(
x
)
=
tan
(
6
x
)
f
'
(
x
)
=
d
d
x
tan
(
6
x
)
=
sec
2
(
6
x
)
⋅
d
d
x
(
6
x
)
(chain rule)
=
6
⋅
sec
2
(
6
x
)
=
6
cos
2
(
6
x
)
(since
sec
=
1
cos
)
So that's our first derivative. Now for our second:
g
(
x
)
=
sin
(
2
x
)
g
'
(
x
)
=
d
d
x
sin
(
2
x
)
=
cos
(
2
x
)
⋅
d
d
x
(
2
x
)
(chain rule)
=
2
⋅
cos
(
2
x
)
Now, all we need to do is combine both of them.
lim
x
→
0
tan
(
6
x
)
sin
(
2
x
)
=
lim
x
→
0
(
6
⋅
sec
2
(
6
x
)
2
⋅
cos
(
2
x
)
)
You can bring the number in front of the limit:
=
6
2
lim
x
→
0
(
sec
2
(
6
x
)
cos
(
2
x
)
)
Now we can replace
sec
by
1
cos
=
3
⋅
lim
x
→
0
(
1
cos
2
(
6
x
)
⋅
cos
(
2
x
)
)
Now let's try entering
x
into this formula:
=
3
⋅
(
1
cos
2
(
6
⋅
0
)
⋅
cos
(
2
⋅
0
)
)
=
3
⋅
(
1
cos
2
(
0
)
⋅
cos
(
0
)
)
=
3
⋅
1
(since
cos
(
0
)
=
1
)
=
3
sunny621:
see the question properly,you are answering for another question
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