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lim
x
→
0
log
(
a
+
x
)
−
log
a
x
+
k
lim
x
→
e
log
x
−
1
x
−
e
=
1
\lim_{x \rightarrow 0} \frac{\log(a+x) - \log a}{x} + k \lim_{x \rightarrow e} \frac{\log x - 1}{x - e} = 1
lim
x
→
0
x
l
o
g
(
a
+
x
)
−
l
o
g
a
+
k
lim
x
→
e
x
−
e
l
o
g
x
−
1
=
1
…
Mathematics
Problem
lim
x
→
0
log
(
a
+
x
)
−
log
a
x
+
k
lim
x
→
e
log
x
−
1
x
−
e
=
1
\lim_{x \rightarrow 0} \frac{\log(a+x) - \log a}{x} + k \lim_{x \rightarrow e} \frac{\log x - 1}{x - e} = 1
lim
x
→
0
x
l
o
g
(
a
+
x
)
−
l
o
g
a
+
k
lim
x
→
e
x
−
e
l
o
g
x
−
1
=
1
, then (a)
k
=
e
(
1
−
1
a
)
k = e\left(1 - \frac{1}{a}\right)
k
=
e
(
1
−
a
1
)
(b)
k
=
e
(
1
+
a
)
k = e(1 + a)
k
=
e
(
1
+
a
)
(c)
k
=
e
(
2
−
a
)
k = e(2 - a)
k
=
e
(
2
−
a
)
(d) The equality is not possible
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