`lim_(x->1) ln(x^3)/(x^2-1)` Evaluate the limit, using L’Hôpital’s Rule if necessary.

`lim_(x->1) ln(x^3)/(x^2-1)`

To solve, plug-in x = 1.

`lim_(x->1) ln(x^3)/(x^2-1) = ln(1^3)/(1^2-1) = 0/0`

Since the result is indeterminate, to determine the limit of the function as x approaches 1, apply L'Hopital's Rule. To do so, take the derivative of the numerator and denominator.

`lim_(x->1) ln(x^3)/(x^2-1) = lim_(x->1) ((ln(x^3))')/((x^2-1)') = lim_(x->1) (1/x^3*3x^2)/(2x) =lim_(x->1)(3/x)/(2x) = lim_(x->1)3/(2x^2)`

And, plug-in x = 1.

`= 3/(2*1) = 3/2`

Therefore, `lim_(x->1) ln(x^3)/(x^2-1)=3/2` .

Comments

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