lim x tends to zero then [100.tanx.sinx÷xsquare]

Question

How to solve this step function

Answer 1

????????????????? 100*sine(x) *tan(x) /x /x ???????????

limit as x gotu 0...[sine(x) /x]...gotu 1

tan(x)=sine/kosine, so tan(x) /x gotu 1/kosine(0)...become 1/1...1

so yu hav 100

Answer 2

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