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If `lim_(xrarr oo) (1+(a)/(x)+(b)/(x^2))^(2x)=e^2`, thenA. `a=1,b=2`B. `a=2,b=1`C. `a=1,b in R`D. `a=b=1` |
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Answer» Correct Answer - C We have, `lim_(xto oo) (1+(a)/(x)+(b)/(x^2))^2x` `=e^(x^(lim_(x to oo)((a)/(x)+(b)/(2))xx2x))=e^xlim_(xto oo)(2a+(2b)/(x))=e^2a` `lim_(xto oo) (1+(a)/(x)+(b)/(x^2))^2x=e^2rArr e^2 rArr e^2a= e^2 rArr 2 rArr a=1`. Hence, `a=1 and b in R`. |
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