Question

$\lim_{x \to 0} \frac{tanx - sinx}{x^3} = \frac{1}{2}$
Why can't it be 0 ?
I got it by :
$\lim_{x \to 0} \frac{tanx - sinx}{x^3}$
$\lim_{x \to 0}\frac{tanx}{x^3} - \frac{sinx}{x^3}$
$\lim_{x \to 0}\frac{1}{x^2} (\frac{tanx}{x} - \frac{sinx}{x})$
$\frac{1}{x^2} (1-1) =\frac{1}{x^2} (0) = 0$
What's wrong with this solution.
Quality answer NEEDED,.

Answer 1

Here tan x/x and sinx/x approaches to 1 when x approaches to 0

approaching to 1 - approaching to 1 is approaching to 0

and 1/x^2 , that is 1/approaching to 0

Approaching to 0/approaching to 0

is indeterminant form

Answer 2

Answer: observation and conclusion

Step-by-step explanation: