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Mathematics
Mathematics
Problem
If
f
(
x
)
=
sin
(
e
x
−
2
−
1
)
log
(
x
−
1
)
f(x) = \frac{\sin(e^{x-2} - 1)}{\log(x-1)}
f
(
x
)
=
l
o
g
(
x
−
1
)
s
i
n
(
e
x
−
2
−
1
)
, then
lim
x
→
2
f
(
x
)
\lim_{x \rightarrow 2} f(x)
lim
x
→
2
f
(
x
)
is given by (a) -2 (b) -1 (c) 0 (d) 1
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