What is the equation of the line tangent to # f(x)=1/ln(x^2+x)*(x^2+x) # at # x=-1 #?

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Answer 1 · 2016-06-15T10:20:03

#y = 0#

With #f(x)=(x^2+x)/(log_e(x^2+x))# we have

#lim_{x->-1}f(x) = 0#

and also taking

#m(x) = (df)/(dx) =(1 + 2 x)/Log_e(x + x^2) -(1 + 2 x)/log_e(x + x^2)^2#

we have

#lim_{x->-1}m(x) = 0#

so the tangent line to #f(x)# at #p_0 = {-1,0}# is given by

#y = 0 + 0(x-0) = 0#