Question

In Exercises 2.4.2-2.4.40, find the indicated limits. $$ \lim _{x \rightarrow 1+}\left(\frac{x+1}{x-1}\right)^{\sqrt{x^{2}-1}} $$

Short answer

Answer: The limit is 1.

Step-by-step solution

Rewrite the expression using logarithm properties

We see that the limit is in the form of \(1^\infty\), which is an indeterminate form. To handle this, let's first take the natural logarithm of the expression and apply logarithm properties: $$ \ln\left( \lim _{x \rightarrow 1+}\left(\frac{x+1}{x-1}\right)^{\sqrt{x^{2}-1}} \right) = \lim _{x \rightarrow 1+}\left( \sqrt{x^{2}-1} \cdot \ln\left(\frac{x+1}{x-1}\right) \right) $$ Now the limit is in the form of \(0 \cdot \infty\), but we can rewrite it as a fraction to apply L'Hopital's Rule.

Rewrite the expression as a fraction

Rewrite the limit as a fraction so that we can apply L'Hopital's Rule: $$ \lim _{x \rightarrow 1+}\left( \frac{\ln\left(\frac{x + 1}{x - 1}\right)}{\frac{1}{\sqrt{x^2 - 1}}} \right) $$ Now we have the limit in the form \(\frac{0}{0}\), which is another indeterminate form that allows us to apply L'Hopital's Rule.

Apply L'Hopital's Rule

By applying L'Hopital's Rule, we differentiate both the numerator and the denominator: $$ \lim _{x \rightarrow 1+} \frac{\frac{d}{dx} \ln\left(\frac{x+1}{x-1}\right)}{\frac{d}{dx} \frac{1}{\sqrt{x^2 - 1}}} $$

Differentiate the numerator and denominator

Differentiate both the numerator and the denominator: $$ \frac{d}{dx} \ln\left(\frac{x+1}{x-1}\right) = \frac{2}{(x-1)(x+1)} $$ $$ \frac{d}{dx} \frac{1}{\sqrt{x^2 - 1}} = -\frac{x}{(x^2 - 1)^{3/2}} $$ Now substitute these derivatives back into the limit expression: $$ \lim _{x \rightarrow 1+} \frac{\frac{2}{(x-1)(x+1)}}{-\frac{x}{(x^2 - 1)^{3/2}}} $$

Rationalize the limit expression

Simplify the limit expression by multiplying the numerator and denominator by the reciprocal of the denominator: $$ \lim _{x \rightarrow 1+} \frac{-2(x^2 - 1)^{3/2}}{x(x - 1)(x + 1)} $$ Now, we can see that the limit evaluates to a definite form.

Evaluate the limit

Evaluate the limit as \(x\) approaches \(1+\): $$ \lim _{x \rightarrow 1+} \frac{-2(x^2 - 1)^{3/2}}{x(x - 1)(x + 1)} = \frac{-2 \cdot 0^{3/2}}{1 \cdot 0 \cdot 2} = 0 $$

Find the original limit

Now that we found the limit of the natural logarithm of the original expression, we can find the original limit by taking the exponential of both sides: $$ \lim _{x \rightarrow 1+}\left(\frac{x+1}{x-1}\right)^{\sqrt{x^{2}-1}} = e^0 = 1 $$ Therefore, the limit of the given expression is 1.

Key concepts

Indeterminate Forms

When we encounter a limit in calculus that presents an outcome like \(0/0\), \(\infty/\infty\), \(0 \cdot \infty\), \(1^\infty\), \(\infty - \infty\), \(0^0\), or \(\infty^0\), these are all examples of indeterminate forms. These forms don’t provide enough information for us to determine the limit's value directly. Hence, additional techniques, like algebraic manipulation and L'Hopital's Rule, become necessary to evaluate them.

For instance, in the given exercise, the limit \(\lim _{x \rightarrow 1+}\left(\frac{x+1}{x-1}\right)^{\sqrt{x^{2}-1}}\) initially appears to approach the form \(1^\infty\), which is indeterminate. This ambiguity occurs because, while 1 raised to any power is generally 1, infinity isn't a concrete number, and thus the limit could potentially diverge or converge to a value that isn't immediately obvious.

L'Hopital's Rule

When faced with indeterminate forms, one powerful tool at our disposal is L'Hopital's Rule. This rule states that if we have a limit of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), we can take the derivative of the numerator and the denominator separately and then take the limit again. If this new limit exists or reaches an infinity, it is the limit of the original expression.

In our example, after first manipulating the limit into a form amenable to L'Hopital's Rule, we differentiate the numerator and the denominator. This is often sufficient to resolve the indeterminacy and find the limit's value. Remember that L'Hopital's Rule can sometimes be applied more than once, and one should check for continuity and differentiability of the functions involved before applying the rule.

Natural Logarithm Properties

The natural logarithm, denoted as \(\ln\), is crucial in simplifying complex expressions and solving limits involving indeterminate forms. One particular property of logarithms that's frequently used is the power rule: \(\ln(a^b) = b \cdot \ln(a)\), where \(a > 0\) and \(b\) are real numbers. This property allows us to transform multiplicative relationships into additive ones, which are often easier to handle.

In our textbook exercise, we utilize this property to simplify the limit by taking the natural logarithm of the entire expression, which helps us transition from a \(1^\infty\) indeterminate form to a \(0 \cdot \infty\) form. From there, we can proceed to manipulate the expression further and apply differentiation techniques.

Differentiation Techniques

To apply L'Hopital's Rule effectively, one must be skilled in differentiation techniques. Differentiation is the process of finding the derivative, which represents the rate at which a function is changing at any given point. Common techniques include the product rule, quotient rule, chain rule, and the power rule, among others.

Throughout the steps of solving the limit in the original exercise, we differentiated both the numerator and the denominator—a step that’s crucial for applying L'Hopital's Rule. Understanding how these techniques work and being able to apply them correctly are essential for tackling limits, and indeed many other aspects of calculus. Not only does this allow us to handle indeterminate forms, but it also serves to refine our mathematical intuition and problem-solving skills.