Question

Let \(f\) be differentiable on some deleted neighborhood \(N\) of \(x_{0}\), and suppose that \(f\) and \(f^{\prime}\) have no zeros in \(N\). Find (a) \(\lim _{x \rightarrow x_{0}}|f(x)|^{f(x)}\) if \(\lim _{x \rightarrow x_{0}} f(x)=0\); (b) \(\lim _{x \rightarrow x_{0}}|f(x)|^{1 /(f(x)-1)}\) if \(\lim _{x \rightarrow x_{0}} f(x)=1 ;\) (c) \(\lim _{x \rightarrow x_{0}}|f(x)|^{1 / f(x)}\) if \(\lim _{x \rightarrow x_{0}} f(x)=\infty\).

Short answer

Question: Find the limits of the following expressions as \(x \rightarrow x_0\), where \(f(x)\) is a differentiable function such that \(\lim_{x \rightarrow x_0} f(x) = 0\) and \(f(x)\) and \(f'(x)\) have no zeros in some deleted neighborhood \(N\) of \(x_0\): (a) \(\lim_{x \rightarrow x_0} |f(x)|^{f(x)}\) (b) \(\lim_{x \rightarrow x_0} |f(x)|^{\frac{1}{(f(x) - 1)}}\) (c) \(\lim_{x \rightarrow x_0} |f(x)|^{\frac{1}{f(x)}}\) Answer: (a) \(\lim_{x \rightarrow x_0} |f(x)|^{f(x)} = 1\) (b) \(\lim_{x \rightarrow x_0} |f(x)|^{\frac{1}{(f(x) - 1)}} = 1\) (c) \(\lim_{x \rightarrow x_0} |f(x)|^{\frac{1}{f(x)}} = 1\)

Step-by-step solution

Rewrite the expression

We'll utilize the fact that \(a^b = e^{b\ln{a}}\) to rewrite the given expression as: $$\lim_{x \rightarrow x_0} |f(x)|^{f(x)} = \lim_{x \rightarrow x_0} e^{f(x)\ln{|f(x)|}}.$$

Apply L'Hopital's rule

To evaluate the limit, we'll use L'Hopital's Rule on the exponent of the expression: $$\lim_{x \rightarrow x_0} \frac{f(x)\ln{|f(x)|}}{1}.$$ Since both the numerator and denominator tend to \(0\) as \(x\rightarrow x_0\), applying L'Hopital's Rule yields: $$\lim_{x \rightarrow x_0} \frac{f'(x)\ln{|f(x)|} + f'(x)\frac{f(x)}{|f(x)|}}{0}.$$ As the denominator is \(0\), we can rewrite this as: $$\lim_{x \rightarrow x_0} \frac{f'(x)\ln{|f(x)|}}{-f'(x)}.$$

Simplify and evaluate the limit

The expression simplifies to: $$\lim_{x\rightarrow x_0} \ln{|f(x)|}.$$ And since \(\lim_{x \rightarrow x_0}f(x)=0\), the limit evaluates to: $$\ln{|\lim_{x \rightarrow x_0} f(x)|} = \ln{0}.$$

Final answer for case a

Now, using the fact that \(a^b = e^{b\ln{a}}\), the final answer for case a is: $$\lim_{x \rightarrow x_0} |f(x)|^{f(x)} = e^{\ln{(0)}} = 1.$$ #Case b and c# Similar to the steps in case a, you can find the limits for cases b and c:

Final answer for case b

In this case, we have: $$\lim_{x \rightarrow x_0} |f(x)|^{\frac{1}{(f(x) - 1)}} = 1.$$

Final answer for case c

In this case, we have: $$\lim_{x \rightarrow x_0} |f(x)|^{\frac{1}{f(x)}} = 1.$$

Key concepts

L'Hôpital's Rule

L'Hôpital's Rule provides a method to evaluate limits that result in indeterminate forms such as \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), and others. It helps find the limit of a ratio of functions by examining the derivatives of the functions involved.
To apply L'Hôpital's Rule effectively:

• Verify that both the numerator and the denominator approach zero or both approach infinity as \(x\) approaches the point of interest.
• Take the derivative of the numerator and the derivative of the denominator.
• Evaluate the new limit of the ratio of these derivatives.

If the limit is still indeterminate, L'Hôpital's Rule can be applied repeatedly as needed until a determinate form is obtained. This rule is particularly useful for problems involving exponential, logarithmic, and polynomial functions that can turn messy when finding limits directly.
Remember, L'Hôpital's Rule only applies when forms like \(\frac{0}{0}\) are present, and using derivatives to simplify expressions can reveal simpler forms to evaluate.

Differentiability

Differentiability means that a function has a derivative at each point of a domain. When we say a function is differentiable around a point \(x_0\), it implies the function is smooth and continuous at \(x_0\), and does not have any sharp corners or cusps at that point.
In the exercise, we're dealing with a differentiable function \(f(x)\), meaning that:

• Both \(f(x)\) and its derivative \(f'(x)\) are well-defined around \(x_0\).
• There are no interruptions or jumps in the graph of \(f(x)\) near \(x_0\).
• The slopes derived from the tangent lines of \(f(x)\) at points around \(x_0\) form a consistent pattern, which is expressed by its derivative \(f'(x)\).

Since differentiability implies continuity, it is a stronger condition than continuity alone. It's important to ensure differentiability when using calculus tools like L'Hôpital's Rule, as manipulating derivatives presupposes the derivative itself is well-behaved.

Exponential Functions

Exponential functions have the form \(a^b\), and they are computed using natural exponentials due to their complex behavior in limits. A crucial identity when dealing with exponential functions in limits applications is \(a^b = e^{b\ln{a}}\).
This transformation is beneficial because:

• Natural exponential \(e\) is smooth and differentiable, making it well-suited for limit evaluations.
• It allows expressing powers as products involving logarithms, which are easier to handle analytically.
• The logarithm \(\ln{a}\) breaks down complex expressions into simpler forms, especially when limits approach 0, 1, or infinity.

In exercises involving exponents, the expression \(e^{b\ln{a}}\) helps convert limits of seemingly complicated powers into limits of simpler functions that are easier to verify and solve using existing calculus tools like L'Hôpital's Rule. Understanding how exponential behavior affects limits and using properties of \(e\) transforms complex problems into manageable calculable tasks.