Question

Lim x---->0 (Sinx -x )/( tan x - x ). How will I solve this limit?

Answer 1

`-oo`

Explanation:

Recall `sin(-x)=-sin(x)`

Therefore we have `lim_(x->0)-sinx/(tanx-x)`

Substitution yields `-sin0/(tan0-0)` which simplifies to `-0/0`

This is indeed indeterminate so we can use L'Hopital's Rule which states

`lim_(x->0)-sinx/(tanx-x)=lim_(x->0) (d/dx(-sinx))/(d/dx(tanx-x)`

This result in `lim_(x->0) -cosx/(sec^2x-1)`

Substitution gives `-cos0/(sec^2(0)-1)`

This results in `-1/(1-1) rarr -1/0` this tends toward `-oo :.`

Answer 2

`lim_(xto0)(sinx-x)/(tanx-x)=-1/2`

Explanation:

Let,

`L=lim_(xto0)(sinx-x)/(tanx-x)=(0/0)` Form

Applying L'Hospital's Rule ,Diff.w.r.t. `x`

`L=lim_(xto0)(cosx-1)/(sec^2x-1)=(0/0)` Form

Again a pplying L'Hospital's Rule:

`L=lim_(xto0)(-sinx)/(2secx*secxtanx`

`L=1/2lim_(xto0)(-sinx)/(sec^2x*sinx/cosx`

`L=1/2lim_(xto0)(-cosx)/sec^2x`

`L=1/2(-cos0)/sec^2 0`

`L=1/2((-1)/1)`

`L=-1/2`