Can someone help me understand how the solution for the following limit is 1/4?
I've been trying to solve it but I always end up in a 'dead end' with an
indetermination. If someone could help me, that would be awesome.
limx→21x−2⋅sin(x−2x+2)
Answer
Note that for x→2
1x−2⋅sin(x−2x+2)=sin(x−2x+2)x−2=sin(x−2x+2)x−2x+2⋅1x+2→1⋅14=14
indeed
y=x−2x+2→0⟹sin(x−2x+2)x−2x+2=sinyy→1
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