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

2 Answers
Jun 26, 2018

-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 :.

Jun 26, 2018

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