Question

limit x tends to zero ([100x/sinx] +[99sinx/x])
where [.] is greatest integer function.

Answer 1

Please see the attachment

Answer 2

The limit x tends to zero ([100x/sinx] +[99sinx/x]) is 198 where [.] is greatest integer function.

• When x tends to 0 x is less than sinx, that is x < sinx
• Hence $\frac{x}{sinx}$ < 1 and $\frac{sinx}{x}$ > 1.

• Hence $100\frac{x}{sinx} < 100$ and $99\frac{sinx}{x} > 99$
• now applying the greatest integer function on the first one we get ,[100x/sinx] = 99 ( As it is strictly less than 100)
• Again applying the greatest integer function on the second one we get, [99sinx/x] = 99 ( As it is strictly greater than 99)
• Hence  limit x tends to zero ([100x/sinx] +[99sinx/x]) = 99 + 99 = 198, where [.] is greatest integer function.
• So the required limit is 198.