lim x-pi/2 cotx/tanx
Answers
Answer:
I checked wolframalpha and it exists. So let's figure out what it is.
y
=
lim
x
→
π
2
(
1
+
2
cot
x
)
tan
x
Generally when you have complicated exponents like this, it's a pretty good idea to use
ln
for the interesting property that exponents can be moved out of the argument and made into coefficients.
ln
y
=
ln
(
lim
x
→
π
2
(
1
+
2
cot
x
)
tan
x
)
A useful property of limits is that logarithms can be brought into limits.
ln
y
=
lim
x
→
π
2
ln
(
(
1
+
2
cot
x
)
tan
x
)
Now we can bring that
tan
x
down as a coefficient, and then reciprocate it (
tan
x
→
1
cot
x
).
ln
y
=
lim
x
→
π
2
tan
x
ln
(
1
+
2
cot
x
)
ln
y
=
lim
x
→
π
2
ln
(
1
+
2
cot
x
)
cot
x
Now, using L'Hopital's rule, we can determine the limit by differentiating the numerator and denominator independently.
ln
y
=
lim
x
→
π
2
1
1
+
2
cot
x
⋅
−
2
csc
2
x
−
csc
2
x
ln
y
=
lim
x
→
π
2
2
1
+
2
cot
x
At this point we can solve this by plugging in regular numbers.
cot
x
=
cos
x
sin
x
Plug in
π
2
to get:
cot
(
π
2
)
=
cos
(
π
2
)
sin
(
π
2
)
=
0
1
=
0
Hence:
ln
y
=
lim
x
→
π
2
2
=
2
But remember, we modified the original function using
ln
. To finish this up, simply undo the natural logarithm using exponentiation via the operation
e
u
where
u
=
2
Step-by-step explanation: