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.