Warm up problem 20: Limits on trigo

Well, you could use l'hopital's... Since csc (0) approaches $\infty$ and tan (0) approaches 0,

But it's much simplier to use the identity csc x tan x= sec x. And sec(0)=1.

We can solve it in this way :- $lim_{x\rightarrow0}cosecx.tanx$ = $lim_{x\rightarrow0}cosecx.\dfrac{sinx}{cosx}$ $lim_{x\rightarrow0}\dfrac{1}{sinx}\dfrac{sinx}{cosx}$ Cancelling sin x we get $lim_{x\rightarrow0}\dfrac{1}{cosx}$ $lim_{x\rightarrow0}secx$ now putting $x=0$ we get, $x=1$

as x->0 sinx->x and tanx->x therefore csc x.tan x=tan x/sin x=x/x=1

cscxtanx=secx x=0; sec(0)= 1

cosecx X tanx =1/ cosx ( cos0=1 ) so ans=1

since csc x tan x = 1/cos x and cos x = 1 as x approaches 0, so we'll get 1/1 =1