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Answer 1 · 2018-01-04T15:47:03

#2#

Using L'Hospital's Rule:

Differentiate numerator and denominator, until the denominator can be expressed as a constant:

#d/dx(e^(2x)-1)=2e^(2x)#

#d/dx(ln(x+1))=1/(x+1)#

#:.#

#(2e^(2x))/(1/(x+1))=(2xe^(2x)+2e^(2x))/1=2xe^(2x)+2e^(2x)#

Plugging in #x=0#

#2(0)e^(2(0))+2e^(2(0))=2#

#:.#

#lim_(x->0)(e^(2x)-1)/(ln(x+1))=2#

Answer 2 · 2018-01-04T17:34:28

#2#

#ln(x + 1)/ (e^(2 x) - 1)= ln((x+1)^(1/(e^(2x)-1)))#but

#e^(2x) = 1+ 2x+(2x)^2/(2!) + cdots +# and for small #absx#

#e^(2x) approx 1+2x# so for small #abs x#

#(x+1)^(1/(e^(2x)-1)) approx (x+1)^(1/(2x))# now

#lim_(x->0) (x+1)^(1/(2x)) = e^(1/2)# then

#lim_(x->0)ln(x + 1)/ (e^(2 x) - 1) = 1/2# and consequently

#lim_(x->0) (e^(2 x) - 1)/ln(x+1) = 2#