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Problems
Let
q
(
x
)
=
{
ln
(
1
+
k
x
)
x
,
x
≠
0
,
2
,
x
=
0.
q(x)= \begin{cases} \displaystyle\frac{\ln(1+kx)}{x}, & x\ne 0,\\ 2, & x=0. \end{cases}
q
(
x
)
=
⎩
⎨
⎧
x
ln
(
1
+
k
x
)
,
2
,
x
=
0
,
x
=
0.
…
Mathematics
Problem
Let
q
(
x
)
=
{
ln
(
1
+
k
x
)
x
,
x
≠
0
,
2
,
x
=
0.
q(x)= \begin{cases} \displaystyle\frac{\ln(1+kx)}{x}, & x\ne 0,\\ 2, & x=0. \end{cases}
q
(
x
)
=
⎩
⎨
⎧
x
ln
(
1
+
k
x
)
,
2
,
x
=
0
,
x
=
0.
For which value of
k
k
k
is
q
q
q
continuous at
x
=
0
x=0
x
=
0
? a)
1
1
1
b)
0
0
0
c)
2
2
2
d)
e
e
e
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