`lim_(x->1) (x/(x - 1) - 1/ln(x))` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t...

`lim_(x->1) (x/(x - 1) - 1/ln(x))` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t apply, explain why.

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Textbook Question

Chapter 4, 4.4 - Problem 49 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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lemjay | High School Teacher | (Level 3) Senior Educator

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`lim_(x->1) (x/(x-1) - 1/(ln(x)))`

To solve, plug-in x=1.

`lim_(x->1) (x/(x-1) - 1/(ln(x))) = 1/(1-1)-1/ln(1)=1/0-1/0=oo-oo`

Since the result is indeterminate, we can apply the L'Hospital's Rule. To do so, subtract the two rational expressions.

`lim_(x->1) (x/(x-1)-1/(ln(x))) = lim_(x->1) ((xln(x))/((x-1)ln(x)) - (x-1)/((x-1)ln(x))) = lim_(x->1) (xln(x) - x + 1)/((x-1)ln(x))` 

Now that we have one rational function, take the derivative of the numerator and denominator.

`lim_(x->1) ((xln(x) - x + 1)')/(((x-1)ln(x))')= lim_(x->1) (x*1/x + 1*ln(x) - 1 + 0)/((x - 1)*1/x + 1*ln(x))= lim_(x->1) (1+lnx-1) /(1-1/x+ln(x))=lim_(x->1) ln(x)/(1-1/x+ln(x))` 

Then, plug-in x = 1.

`= ln(1)/(1-1/1+ln1) = 0/0`

Since the result is still indeterminate, apply the L'Hospital's Rule again. So, take the derivative of the numerator and denominator.

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

Then, plug-in x=1.

`=1/(1+1) = 1/2`

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

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