Determine the behavior of limits

A. Limit as x approaches 1 of:

(log x)/((x-1)^2)

B. Limit as x approaches infinity of:

((x-1)^2)/(log x))

taking derivatives, we have

(1/x) / (2(x-1)) = 1/0 = ±∞

B is the same, since as x->∞,
(x-1)^2 / logx is the same as
1 / (logx / (x-1)^2) as x->1

To determine the behavior of limits, we need to evaluate the limit as the variable approaches a specific value or infinity. Let's solve each problem:

A. Limit as x approaches 1 of (log x)/((x-1)^2):

To evaluate this limit, we can try direct substitution. When x approaches 1, we substitute x = 1 into the expression:

(log 1)/((1 - 1)^2) = (log 1)/0

The denominator becomes 0, which indicates an indeterminate form. We cannot determine the limit through direct substitution.

To further evaluate the limit, we can apply a method called L'Hôpital's Rule. L'Hôpital's Rule states that if we have a limit in an indeterminate form 0/0 or ∞/∞, we can take the derivative of the numerator and the denominator until we obtain a non-indeterminate form.

Taking the derivative of the numerator and denominator, we get:

d/dx (log x) / d/dx (x-1)^2

Using the chain rule, we differentiate log x as 1/x and (x-1)^2 as 2(x-1). Simplifying the expression, we have:

(1/x)/(2(x-1))

Now, evaluate the limit as x approaches 1 again:

lim(x->1) (1/x)/(2(x-1))

Substituting x = 1, we obtain:

lim(x->1) (1/1)/(2(1-1)) = 1/0

This result is still an indeterminate form. We need to apply L'Hôpital's Rule once again.

Differentiating the numerator and denominator:

d/dx (1/x) / d/dx (2(x-1))

The derivative of (1/x) is -1/x^2 and the derivative of 2(x-1) is 2.

Simplifying the expression, we get:

(-1/x^2)/2

Now, evaluate the limit as x approaches 1 again:

lim(x->1) (-1/x^2)/2

Substituting x = 1:

lim(x->1) (-1/1^2)/2 = -1/2

Therefore, the limit as x approaches 1 of (log x)/((x-1)^2) is -1/2.

B. Limit as x approaches infinity of ((x-1)^2)/(log x):

In this case, as x approaches infinity, the logarithm term dominates the expression.

When x becomes very large, we can approximate the behavior of the expression by assuming log x approaches infinity:

lim(x->∞) ((x-1)^2)/(log x)

As log x approaches infinity, the denominator becomes significantly larger than the numerator.

Since we have a finite numerator and an increasing denominator, the limit in this case will be 0.

Therefore, the limit as x approaches infinity of ((x-1)^2)/(log x) is 0.