kpham
evaluate: lim x>2 sqrt (6x)2/sqrt (3x)1



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kpham
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valuate: lim x>2 [sqrt (6x)]2/[sqrt (3x)]1

kpham
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help me plzz

zzr0ck3r
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do u know lhopital rule?

kpham
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whats that

kpham
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im learning about limits and valuate limit

rainbow22
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Square top and bottom?

kpham
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top is sqrt of 6x only
bottom is sqrt of 3x only

febylailani
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two sided limit doesnt exist.
