Evaluate the following limit.
limx→0+(sin x)x
What i have tried:
ln [limx→0+(sin x)x]
limx→0+ln (sin x)x
limx→0+ln (sin x)1x
Applying l'hopital's rule.
limx→0+cot x−x−2
If i keep applying l'hopital's rule, i get indeterminate form. Is what Iam doing right ?
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