Lim x tends to 0 xtanx/sin3x
and
lim x tends to infty 1/n {(m+1)(m+2) ... (m+n)}^1/n
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ae tum kon ho
Step-by-step explanation:
Lim x tends to 0 xtanx/sin3x
and
lim x tends to infty 1/n {(m+1)(m+2) ... (m+n)}^1/n
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