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Chapter 24: Problem 3

Show that \(\lim _{x \rightarrow 0} \frac{\operatorname{erf} x}{x}=\frac{2}{\sqrt{\pi}}\).

Short Answer

Expert verified
The limit is \( \frac{2}{\sqrt{\pi}} \).

Step by step solution

01

Understanding the Error Function

The error function, denoted as \( \operatorname{erf}(x) \), is given by the integral \( \operatorname{erf}(x) = \frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{-t^2} \, dt \). Our task is to find the limit \( \lim _{x \rightarrow 0} \frac{\operatorname{erf} x}{x} \).
02

Apply L'Hopital's Rule

Since both the numerator \( \operatorname{erf}(x) \) and the denominator \( x \) approach 0 as \( x \rightarrow 0 \), the expression has an indeterminate form \( \frac{0}{0} \). Thus, we apply L'Hopital's Rule, which involves differentiating the numerator and the denominator.
03

Differentiate the Numerator and Denominator

The derivative of the numerator \( \operatorname{erf}(x) \) is \( \frac{d}{dx} \left(\frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{-t^2} \, dt \right) = \frac{2}{\sqrt{\pi}} e^{-x^2} \). The derivative of the denominator \( x \) is 1.
04

Compute the Limit Using Differentiated Form

By applying L'Hopital's Rule, we get the limit:\[\lim _{x \rightarrow 0} \frac{\operatorname{erf} x}{x} = \lim _{x \rightarrow 0} \frac{\frac{2}{\sqrt{\pi}} e^{-x^2}}{1} = \frac{2}{\sqrt{\pi}}\]As \( x \) approaches 0, \( e^{-x^2} \) approaches \( e^0 = 1 \), leading to the result \( \frac{2}{\sqrt{\pi}} \).
05

Conclusion

The limit \( \lim _{x \rightarrow 0} \frac{\operatorname{erf} x}{x} = \frac{2}{\sqrt{\pi}} \) is confirmed by applying derivatives and evaluating the resulting expression with L'Hopital's Rule.

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Key Concepts

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

Error Function
The error function, denoted as \( \operatorname{erf}(x) \), is a special mathematical function with a broad range of applications, especially in probability, statistics, and partial differential equations. Its definition involves an integral:\[\operatorname{erf}(x) = \frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{-t^2} \, dt\]
  • The \( \operatorname{erf}(x) \) function computes the probability that a random variable with a normal distribution will fall between the mean and \( x \) standard deviations.
  • A critical point to remember is that, as \( x \) approaches zero, the values of the error function \( \operatorname{erf}(x) \) also approach zero. This property is crucial for solving limits involving the error function.
Understanding this error function is crucial, as it helps in determining various probabilities and has extensive allusions towards Gaussian functions, which commonly appear in fields like engineering and physics.
L'Hopital's Rule
L'Hopital's Rule is a powerful mathematical tool used to evaluate limits that result in indeterminate forms, such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When faced with these forms, L'Hopital's Rule states that:\[\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}\]provided the limit on the right-hand side exists. In simpler terms, the rule allows us to take the derivative of the numerator and the denominator separately to find the limit.
  • L'Hopital's Rule significantly simplifies the evaluation of limits, especially when the initial form presents complexities due to its indeterminate nature.
  • Remember that the differentiability of both the numerator and denominator is a necessity for applying this rule effectively.
      • In our exercise, using L'Hopital's Rule transforms the indeterminate form \( \frac{0}{0} \) into a more approachable form by using derivatives, enabling us to solve the limit of the error function.
Indeterminate Form
In calculus, an indeterminate form occurs when a limit cannot be directly determined from the initial terms presented. Forms like \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), and \( 0 \times \infty \) suggest that further analysis is needed because they do not immediately provide information about the true behavior of the function.
  • Indeterminate forms require additional techniques, like L'Hopital's Rule or algebraic manipulation, to evaluate the limit correctly.
  • These forms mark the starting point for exploring function behavior at particular points where common intuition about function limits might fail.
In this specific exercise, the limit \( \lim_{x \to 0} \frac{\operatorname{erf}(x)}{x} \) initially gives \( \frac{0}{0} \) as \( x \) approaches zero, highlighting the necessity to apply L'Hopital's Rule. By doing so, the derivatives provide a pathway to solve the limit problem effectively.

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Most popular questions from this chapter

Derive the result $$A(s)=L\left\\{t^{-3 / 2} \exp \left(-\frac{k^{2}}{t}\right)\right\\}=\frac{\sqrt{\pi}}{k} \exp (-2 k \sqrt{s}), \quad k>0, s>0$$ directly from the definition of a transform. In the integral $$A(s)=\int_{0}^{\infty} \exp \left(-s t-k^{2} t^{-1}\right) t^{-3 / 2} d t$$ put \(\beta=\sqrt{t}\) to get $$A(s)=2 \int_{0}^{\infty} \beta^{-2} \exp \left(-s \beta^{2}-k^{2} \beta^{-2}\right) d \beta$$ Or $$A(s)=2 \exp (-2 k \sqrt{s}) \int_{0}^{\infty} \beta^{-2} \exp \left[-\left(\beta \sqrt{s}-k \beta^{-1}\right)^{2}\right] d \beta$$ Show that $$\begin{aligned}\frac{d A}{d s} &=-2 \int_{0}^{\infty} \exp \left(-s \beta^{2}-k^{2} \beta^{-2}\right) d \beta \\\&=-2 \exp (-2 k \sqrt{s}) \int_{0}^{\infty} \exp \left[-\left(\beta \sqrt{s}-k \beta^{-1}\right)^{2}\right] d \beta\end{aligned}$$ Thus arrive at the differential equation $$\sqrt{s} \frac{d A}{d s}-k A=-2 \sqrt{\pi} \exp (-2 k \sqrt{s})$$ and from it obtain the desired function \(A(s)\).

Let \(c>0, s>0,\) and let \(L^{-1}\\{f(s)\\}=F(t) .\) Prove that $$ L^{-1}\\{f(s) \tanh (c s)\\}=F(t)+2 \sum_{n=1}^{\infty}(-1)^{n} F(t-2 n c) \alpha(t-2 n c) $$

Use integration by parts to show that $$\int_{0}^{x} \operatorname{erf} y d y=x \operatorname{erf} x-\frac{1}{\sqrt{\pi}}\left[1-\exp \left(-x^{2}\right)\right]$$.

Show that erf \(x\) is an odd function of \(x\).

Show that for all real \(x,|\operatorname{erf} x|<1\).
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