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

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Answer 1 · 2018-06-26T04:28:48

#-oo#

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 · 2018-06-26T04:30:04

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

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#