How do you find #lim_(x->0^+)(x+sinx)^(tanx)# ?

1 Answer
Gregorio Manea
Feb 19, 2018

#lim_(x->0^+)(x+sinx)^tanx=1#

Explanation:

The function #f(x)=(x+sinx)^tanx# with #f:(0,oo)->(0,oo)# has the following graphic:
graph{y=(x+sinx)^tanx [-0.282, 4.718, -0.34, 2.16]}
#lim_(x->0,x>0)(x+sinx)^tanx=#
#=lim_(x->0,x>0)(x(1+sinx/x))^(sinx/cosx)=#
#=lim_(x->0,x>0)(x^sinx(1+sinx/x)^sinx)^(1/cosx)= #

#lim_(x->0)x^sinx=1#
#lim_(x->0)sinx/x=1#
#lim_(x->0)sinx =0#
#lim_(x->0)cosx=1#

#=(1*(1+1)^0)^(1/1)= (1*1)^1=1#

Related questions
Impact of this question
4704 views around the world
You can reuse this answer
Creative Commons License