Question

Lim_x→0 (sin x -x) / tan x

Answer 1

We need to calculate $lim(x\to0) \frac {(sinx-x)} {tanx}$

It is in $\frac 0 0$ form after putting in place of $x$ .So L Hospital's rule is applicable here .

$= lim(x\to0)\frac {\frac {d} {dx} (sinx-x)} {\frac {d} {dx}( tanx)}$

$=lim(x\to0)\frac {(cosx-1)} {(sec^2x)}$

$=\frac {(cos0 -1)} {(sec^2(0)}$

$=\frac {(1-1)} {1^2}$ ( As $cos(0)=1, sec(0)=1)$

=$\frac 0 1$

=0 (Ans)