Question: lim x tends to zero then [100.tanx.sinx÷xsquare].
Use l'Hôpital's rule here.
Let f(x) = g(x)/h(x),
where
g(x)=100(tanx.sinx)
h(x) = x^2
Then Lim [x->0] f(x) = Lim [x->0] g'/h'
g' = 100(1+sec^2(x)).sin(x)
h' = 2x
But g'(0)/h'(0) is indeterminate, so use l'Hôpital's rule once more.
g'' = 100.(cos(x)^4-cos(x)^2+2)/cos(x)^3
h'' = 2
This time g''(0) = 100.(1 - 1 + 2)/1 = 200
and h''(0) = 2
So, Lim [x->0] f = Lim [x->0] g''/h'' = 200/2 = 100
Answer: Limit = 100