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Chapter 4: Problem 61

\(5-64\) Find the limit. Use I'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If l'Hospital's Rule doesn't apply, explain why. $$\lim _{x \rightarrow 0^{+}}(4 x+1)^{\cot x}$$

Short Answer

Expert verified
The limit is \( e^4 \).

Step by step solution

01

Recognize the Form of the Limit

The given limit is \( \lim_{x \rightarrow 0^{+}}(4x+1)^{\cot x} \). As \( x \rightarrow 0^+ \), \( 4x+1 \rightarrow 1 \) and \( \cot x \rightarrow \infty \), making this a \(1^\infty\) indeterminate form. This suggests the use of the exponential limit theorem.
02

Apply the Exponential Limit Theorem

For an expression \((a_n)^{b_n}\) where \((a_n)\to 1\) and \(b_n\to\infty\), the limit is \( \exp\left(\lim_{x \to c} (a_n - 1) \cdot b_n \right) \) if the latter limit exists. Set \( L = \lim_{x \to 0^+} \left( (4x + 1)^{\cot x} \right) \). Rewrite it as \( \exp\left( \lim_{x \to 0^+} \cot x \cdot \ln(4x + 1) \right) \).
03

Simplify the Logarithmic Expression

Using the series expansion for \( \ln(4x + 1) \) when \( x \) is close to 0, we have \( \ln(4x + 1) \approx 4x \). So, \( \lim_{x \to 0^+} \cot x \cdot \ln(4x + 1) \approx \lim_{x \to 0^+} \cot x \cdot 4x \).
04

Apply L'H么pital's Rule to the Limit

Now compute \( \lim_{x \to 0^+} \cot x \cdot 4x = \lim_{x \to 0^+} \frac{4x}{\tan x} \). This is an \( \frac{0}{0} \) form, so L'H么pital's Rule applies. Differentiate the numerator and denominator to get \( \lim_{x \to 0^+} \frac{4}{\sec^2 x} = \lim_{x \to 0^+} 4 \cos^2 x = 4 \cdot 1^2 = 4 \).
05

Calculate the Final Limit

With \( \lim_{x \to 0^+} \cot x \cdot \ln(4x + 1) = 4 \), it follows that \( L = \exp(4) \). Thus, the original limit evaluates to \( e^4 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Indeterminate Forms
When tackling limits in calculus, sometimes you encounter expressions that do not have well-defined values at first glance. These are called indeterminate forms. One common indeterminate form is the notorious \(1^\infty\) that appears when trying to compute something like \(\lim_{x \rightarrow 0^{+}} (4x+1)^{\cot x}\). Here, as \(x\) approaches zero:
  • \(4x+1\) approaches 1
  • \(\cot x\) approaches infinity
This unique combination results in the indeterminate form of \(1^\infty\), suggesting ambiguity in the straightforward computation of the limit. In these scenarios, clever mathematical methods like the exponential limit theorem and L'H么pital's Rule come into play to rearrange and resolve the indainty, giving us a precise value.
Exponential Limit Theorem
The Exponential Limit Theorem is a handy tool to evaluate limits like \((a_n)^{b_n}\) when \((a_n)\) tends to 1 and \(b_n\) trends towards infinity. It is particularly useful for indeterminate forms of \(1^\infty\). The theorem helps transform the expression into something more manageable:
  • Rewrite as \(\exp(\lim_{x \to c} (a_n - 1) \cdot b_n)\)
  • Evaluate the limit \(\lim_{x \to c} (a_n - 1) \cdot b_n\)
For example, in our exercise, we rewrote \((4x+1)^{\cot x}\) as \(\exp(\lim_{x \to 0^+} \cot x \cdot \ln(4x + 1))\). This step essentially breaks down the complicated power expression into an exponential form that can be further simplified with other techniques.
Series Expansion
The series expansion is a method of approximating complicated expressions into simpler, usually polynomial, forms. When dealing with logarithmic functions such as \(\ln(4x +1)\) as \(x\) approaches zero, the series expansion formula \(\ln(1 + u) \approx u\) for small \(u\) comes in handy. Here:
  • Approximate \(\ln(4x+1) \approx 4x\)
This simplifies our calculations significantly because it's easier to work with a linear approximation \(4x\) under the limit than a full logarithmic expression. This step was crucial, leading us to compute \(\lim_{x \to 0^+} \cot x \cdot \ln(4x + 1) \approx \lim_{x \to 0^+} \cot x \cdot 4x\), a much more tractable form to handle with L'H么pital's Rule.
Differentiation of Trigonometric Functions
Differentiation plays a fundamental role when applying L'H么pital's Rule, especially with trigonometric functions. For a function like \(\cot x\), we need its derivative if we find a \(\frac{0}{0}\) indeterminate form. L'H么pital's Rule helps by stating:
  • Take derivatives of the numerator and the denominator separately.
  • Find the limit of the resulting fractional expression.
In our case, the challenge is to differentiate the expression \(\frac{4x}{\tan x}\). We differentiate using simple rules:
  • Derivative of \(4x\) is \(4\).
  • Derivative of \(\tan x\) is \(\sec^2 x\).
Thus, applying L'H么pital's Rule simplifies \(\frac{4x}{\tan x}\) to \(\frac{4}{\sec^2 x}\), allowing us to compute \(\lim_{x \to 0^+} 4 \cos^2 x = 4\) easily. This confirms the limit, thus resolving the original challenge.

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Most popular questions from this chapter

Describe how the graph of \(f\) varies as \(c\) varies. Graph several members of the family to illustrate the trends that you discover. In particular, you should investigate how maximum and minimum points and inflection points move when c changes. You should also identify any transitional values of \(c\) at which the basic shape of the curve changes. \(f(x)=\ln \left(x^{2}+c\right)\)

\(5-8\) Use Newton's method with the specified initial approxima- tion \(x_{1}\) to find \(x_{3},\) the third approximation to the root of the given equation. (Give your answer to four decimal places.) \(\frac{1}{3} x^{3}+\frac{1}{2} x^{2}+3=0, \quad x_{1}=-3\)

\(57-62\) A particle is moving with the given data. Find the position of the particle. \(a(t)=t^{2}-4 t+6, \quad s(0)=0, \quad s(1)=20\)

Use Newton's method with initial approximation \(x_{1}=1\) to find \(x_{2},\) the second approximation to the root of the equa- tion \(x^{4}-x-1=0 .\) Explain how the method works by first graphing the function and its tangent line at ( \(1,-1 )\) .

(a) Use Newton's method with \(\mathrm{x}_{1}=1\) to find the root of the equation \(\mathrm{x}^{3}-\mathrm{x}=1\) correct to six decimal places. b) Solve the equation in part (a) using \(x_{1}=0.6\) as the initial approximation. (c) Solve the equation in part (a) using \(x_{1}=0.57 .\) (You defi- nitely need a programmable calculator for this part.) (d) Graph \(f(x)=x^{3}-x-1\) and its tangent lines at \(x_{1}=1\) \(0.6,\) and 0.57 to explain why Newton's method is so sen- sitive to the value of the initial approximation.v
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