How do you find the limit of # ( x + sin(x) ) / (x) # as x approaches 0?

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Answer 1 · 2018-05-22T16:58:33

# 2#.

Knowing that, #lim_(x to 0)sinx/x=1#, we have,

#lim_(x to 0)(x+sinx)/x#,

#=lim_(x to 0){x/x+sinx/x}#,

#=lim_(x to 0){1+sinx/x}#,

#=1+1#

#=2#.

Answer 2 · 2018-05-22T17:07:55

#color(blue)[lim_(xrarr0)x/x+lim_(xrarr0)sinx/x=1+1=2]#

Note that:

#color(red)[lim_(nrarr0)sinn/n=1#

#lim_(xrarr0)( x + sin(x) ) / (x)#

#lim_(xrarr0)x/x+lim_(xrarr0)sinx/x#

#lim_(xrarr0)x/x=0/0#

since the direct compensation product equal #0/0#
we will use L'hospital Rule.

L'hospital Rule #color(red)[lim_(trarra)(f'(x))/(g('x))]#

#lim_(xrarr0)x/x=lim_(xrarr0)1/1=1#

#lim_(xrarr0)sinx/x=1#

#color(blue)[lim_(xrarr0)x/x+lim_(xrarr0)sinx/x=1+1=2]#