#lim_(x->0)tan(x)/(x+sin(x))#?

2 Answers
Eddie
Mar 14, 2017

#= 1/2#

Explanation:

#lim_(x->0)tan(x)/(x+sin(x))#

Using Taylor Expansions:

#=lim_(x->0)(x + x^3/3 + mathcal O x^5)/(x+(x - x^3/(3!) + mathcal O x^5))#

#=lim_(x->0)(x + x^3/3 + mathcal O x^5)/(2x- x^3/(3!) + mathcal O x^5)#

#=lim_(x->0)(1 + x^2/3 + mathcal O x^4)/(2- x^2/(3!) + mathcal O x^4)#

#=1/2#

Jim H
Mar 14, 2017

#tanx/(x+sinx) = (sinx/cosx)/(x+sinx)#

# = sinx/(cosx(x+sinx))#

# = (sinx/x)/(cosx(1+sinx/x))# for all #x != 0#

Now use #lim_(xrarr0) sinx/x =1# to get

#lim_(xrarr0)tanx/(x+sinx) = lim_(xrarr0) (sinx/x)/(cosx(1+sinx/x))#

# = 1/(x(1+1)) = 1/2#

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