How do you find the limit of #[x - ln(x^2+1)]# as x approaches infinity?

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Answer 1 · 2017-02-05T01:53:58

#lim_(x->oo) x-ln(x^2+1) =+oo#

As #x =ln e^x#, we can write the function as:

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

Using the properties of logarithms then we have:

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

Now we know the the exponential is an infinite of higher order than any polynomial for #x->oo#, so

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

and then as the logarithm is continuous in #(0,+oo)#

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