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

Question

Isabella Ackerman · Accepted Answer

This limit 'creates' the #infty - infty# indeterminate form so the first step should be finding a common denominator.

#lim_(x to 1) (1/ln(x)-1/(x-1))=lim_(x to 1) (x-1-ln(x))/(ln(x)(x-1))=[0/0]# And now to get rid of #0/0# you can use the de L'Hôspital's Rule which states that when evaluating #0/0# or #infty/infty# indeterminate forms the limit of the quotient stays the same if derivatives of the numerator and denominator (evaluated seperately, not using the formula for the derivative of a quotient) are substituted for the numerator and denominator, respectively (as long as the 'new' limit exists). So (in this case using the de L'Hôpital's Rule twice): #lim_(x to 1) (1-1/x)/((x-1)/x +ln(x))=[0/0]=lim_(x to 1) (1/x^2)/(1/x^2 +1/x)=[1/(1+1)]=1/2#

Christopher Allers · Answer

undefined To find the limit of (1/(ln x) - 1/(x-1)) as x approaches 1, we can use algebraic manipulation and L'Hôpital's rule.

First, let's simplify the expression by finding a common denominator:

(1/(ln x) - 1/(x-1)) = ((x-1)/(ln x(x-1))) - (ln x/(ln x(x-1)))

Next, combine the fractions:

((x-1) - ln x)/(ln x(x-1))

Now, we can apply L'Hôpital's rule by taking the derivative of the numerator and denominator separately:

Numerator: d/dx ((x-1) - ln x) = 1 - 1/x
Denominator: d/dx (ln x(x-1)) = (1/x)(x-1) + ln x

Simplifying the derivatives:

Numerator: 1 - 1/x
Denominator: (x-1)/x + ln x

Now, we can evaluate the limit as x approaches 1 by plugging in the value:

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

Since we still have an indeterminate form, we can apply L'Hôpital's rule again:

Numerator: d/dx (1 - 1/x) = 1/x^2
Denominator: d/dx ((x-1)/x + ln x) = (1/x) - (x-1)/x^2 + 1/x

Simplifying the derivatives:

Numerator: 1/x^2
Denominator: 1/x - (x-1)/x^2 + 1/x

Now, we can evaluate the limit as x approaches 1:

(1/1^2)/(1/1 - (1-1)/1^2 + 1/1) = 1/1 = 1

Therefore, the limit of (1/(ln x) - 1/(x-1)) as x approaches 1 is 1.