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Problems
Let
g
(
x
)
=
{
x
2
e
−
x
,
x
≥
0
log
3
(
1
+
x
)
x
,
−
1
<
x
<
0
g(x)=\begin{cases} x^{2}e^{-x}, & x\ge 0 \\ \dfrac{\log_{3}(1+x)}{x}, & -1<x<0 \end{cases}
g
(
x
)
=
⎩
⎨
⎧
x
2
e
−
x
,
x
lo
g
3
(
1
+
x
)
,
x
≥
0
−
1
<
x
<
0
…
Mathematics
Problem
Let
g
(
x
)
=
{
x
2
e
−
x
,
x
≥
0
log
3
(
1
+
x
)
x
,
−
1
<
x
<
0
g(x)=\begin{cases} x^{2}e^{-x}, & x\ge 0 \\ \dfrac{\log_{3}(1+x)}{x}, & -1<x<0 \end{cases}
g
(
x
)
=
⎩
⎨
⎧
x
2
e
−
x
,
x
lo
g
3
(
1
+
x
)
,
x
≥
0
−
1
<
x
<
0
. Determine
lim
x
→
0
g
(
x
)
\displaystyle\lim_{x\to 0} g(x)
x
→
0
lim
g
(
x
)
. a)
0
0
0
b)
ln
3
\ln 3
ln
3
c) The limit does not exist d)
1
ln
3
\dfrac{1}{\ln 3}
ln
3
1
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