Examine which is greater : sin x tan x or `x^(2)`, Hence evaluate `lim_(hto0) [(sin x tanx)/(x^(2))], " Where " x in (0,(pi)/(2))`

Explanation

`"Let "f(x) sin x tan x =x^(2)`
` f(x) =cos x , tan x +sin x . sec^(2) -2x``rArr f(x) =sin x + sin x sec^(2) -2x`
`rArr f(x) =cos x + cos x sec^(2) +2sec^(2) sin x tan x-2`
`rArr f(x) =(cos x + sec x -2) +2 sec^(2) x sin x tan x`
`"Now " cos x + sec x -2 = (sqrt(cos ) x -sqrt( sec ) x)^(2)` and 2 `sec^(2) x tan x.` sin x `gt` 0 because `x in (0,(pi)/(2))`
`rArr f(x) gt0 rArr f(x) " is " M.I.`
`" Hence " f(x) gt f(0)`
`rArr f(x) gt 0 rArr f(x) " is "M.I rArr f(x) gt0`
`rArr sin x tan x -x^(2) gt0`
`"Hence sin x tan x " gt x^(2) rArr (sin x tan x)/(x^(2)) gt 1 rArr underset(hto0)("lim") [(sin x tanx)/(x^(2))] =1.`