Lim x tending to 0 sin(x+h)log(x+h)-Sind logx/h
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Answer:
lim
h
→
0
ln
(
x
+
h
)
−
ln
x
h
=
1
x
Explanation:
We seek:
L
=
lim
h
→
0
ln
(
x
+
h
)
−
ln
x
h
Method 1:
L
=
lim
h
→
0
ln
(
x
+
h
x
)
h
=
lim
h
→
0
ln
(
1
+
h
x
)
h
Now, we can perform a substitution:
Let
z
=
h
x
and we note that
z
→
0
as
h
→
0
Then, we have:
L
=
lim
z
→
0
ln
(
1
+
z
)
z
x
=
lim
z
→
0
1
x
1
x
ln
(
1
+
z
)
=
1
x
lim
z
→
0
1
z
ln
(
1
+
z
)
=
1
x
lim
z
→
0
ln
(
1
+
z
)
1
z
Du to the monotonicity of the logarithmic function, we can change the limit to:
=
1
x
ln
{
lim
z
→
0
(
1
+
z
)
1
z
}
And we note that this is a standard limit , established by Leonhard Euler :
lim
z
→
0
(
1
+
z
)
1
z
=
e
Giving us:
L
=
1
x
ln
e
=
1
x
Method 2:
If we compare the sought limit:
L
=
lim
h
→
0
ln
(
x
+
h
)
−
ln
x
h
With the limit definition of the derivative:
f
'
(
x
)
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
Then, we note that
L
=
d
d
x
(
ln
x
)
, leading to the same result as above.
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