Evaluate : `[lim_(x to 0) (tan x)/(x)]`, where `[*]` represents the greatest integer function.

Explanation

Let us first draw the graph of y = x and y = tan x.
For `y = x, y' = 1`, i.e. the slope of the line is 1 for all real x.
For `y = tan x , y' = sec^(2)x`
Thus, both y = x and y = tan x have slope '1' at x = 0, hence the graphs touch each other.
Moving from '0' to `'pi//2'` the values of `sec^(2)x` increase from '1' to `oo`.
Thus, the graph pf y = tan x lies below the graph of y = x.
The graphs of y = x and y = tan x are plotted as follows.

From the graph, when `x to 0^(+)`, the graph of y = tan x is above the graph of y = x
or `tan x gt x rArr (tan x )/(x) gt 1 rArr [underset(x to 0^(+))lim (tan x)/(x)] = 1`
When `x to 0^(-)`, the graph of y = x is above the graph of y = tan x
or `tan x lt x rArr (tan x )/(x) gt 1` (as x is negative) `rArr [underset(x to 0^(+))lim (tan x)/(x)] = 1`
Thus, `[underset(x to 0)lim (tan x)/(x)] = 1`