lim（x→0）（（arctan（x∧2））╱（sin（x╱2）sinx））

lim（x→0）（（arctan（x∧2））╱（sin（x╱2）sinx））

In fact, it's easier to use the equivalent infinitesimal method here,
When x goes to zero,
SiNx is equivalent to x, and arctanx is also equivalent to X,
So here arctan (x ^ 2) is equivalent to x ^ 2, sin (x / 2) is equivalent to X / 2
that
Original limit
=lim(x->0) x^2 / (x/2 *x)
= 2
So the limit is 2

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