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The options to the above question are 0 1 ¼ none of the above

The options to the above question are
0
1
¼
none of the above
 

Question Image
Grade:12th pass

1 Answers

Aditya Gupta
2080 Points
2 years ago
Write it as (sinx-cosx*x^2)/x^2sinx
Now use series expansion of sin and cos
(x-x^3/6+...-x^2 (1-x^2/2+....))/x^3*[x/sinx]
After a bit of manipulation you will see that the terms DON'T cancel out and instead we get smth like
(1+x)/x^2 
So the limit above is plus infinite.
 

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