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How do you find the limit of #(ln x^2) / (x^2-1)# as x approaches 1?

Answer 1

#lim_{x->1}(log_e x^2) / (x^2-1)=1#

#(log_e x^2) / (x^2-1) = log_e x/(x-1)-log_e x/(x+1)# but the #log_e x# series expansion in the #x=1# vicinity is #log_e x = sum_{k=1}^{oo}((x-1)^k)/k# so

#log_e x/(x-1) = 1 + sum_{k=2}^{oo}((x-1)^{k-1})/k#

#lim_{x->1}log_e x/(x-1) = 1#

Finally

#lim_{x->1}(log_e x^2) / (x^2-1) = lim_{x->1}log_e x/(x-1)-lim_{x->1}log_e x/(x+1) = 1 + 0=1 #

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Answer 2

Paisley Johnson

To find the limit of (ln x^2) / (x^2-1) as x approaches 1, we can use L'Hôpital's Rule. Taking the derivative of the numerator and denominator separately, we get (2/x) / (2x). Simplifying this expression, we have 1/x^2. Substituting x=1 into this expression, we find that the limit is 1.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.