`lim_(x->1) lnx/sin(pix)` Evaluate the limit, using L’Hôpital’s Rule if necessary.

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

To solve, plug-in x = 1.

`lim_(x->1) (ln(x))/(sin(pix)) = (ln(1))/(sin(pi*1)) = 0/0`

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

`lim_(x->1) (ln(x))/(sin(pix)) =lim_(x->1) ((ln(x))')/((sin(pix))') = lim_(x->1) (1/x)/(pi...

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

To solve, plug-in x = 1.

`lim_(x->1) (ln(x))/(sin(pix)) = (ln(1))/(sin(pi*1)) = 0/0`

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

`lim_(x->1) (ln(x))/(sin(pix)) =lim_(x->1) ((ln(x))')/((sin(pix))') = lim_(x->1) (1/x)/(pi cos(pix)) = lim_(x->1) 1/(pix cos(pix))`

And, plug-in x = 1.

`= 1/(pi*1*cos(pi*1))=1/(pi*cos(pi)) = 1/(pi*(-1)) = -1/(pi)`

 

Therefore,  `lim_(x->1) (ln(x))/(sin(pix))=-1/pi` .

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