Find the limit as x-->0 for y = (e^x- 1)/[sin(nx)]
Find the limit as x-> 0 for y = (e^x- 1)/[sin(nx)]
a. Using L’Hopital’s rule
b. Using the series approximation of e^u = 1 + u and sin(u) = u for u << 1
Jayellis00
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Aman R
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The answer is accepted.
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