lim x tends to zero (1-cosmx)/1-cosnx
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Using L'Hopital Rule, the limit becomes,
m sin mx/n sin nx
Using L'Hopital Rule again,
m² cos mx/n² cos nx=m²/n² when x tends to 0
m sin mx/n sin nx
Using L'Hopital Rule again,
m² cos mx/n² cos nx=m²/n² when x tends to 0
sohith8:
thank u very much
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