Problem 36 - Thomas Calculus

Question

Use series to evaluate the limits. \begin{equation} \lim _{x \rightarrow \infty}(x+1) \sin \frac{1}{x+1} \end{equation}

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Rewrite the expression using substitution

Let \( y = \frac{1}{x+1} \), which implies \( x+1 = \frac{1}{y} \). When \( x \to \infty \), \( y \to 0^+ \). Thus, the expression \( (x+1) \sin\left(\frac{1}{x+1}\right) \) becomes \( \frac{1}{y} \sin(y) \).

Step 2: Use the Taylor series expansion for sin(y)

The Taylor series expansion for \( \sin(y) \) around \( y = 0 \) is \( \sin(y) = y - \frac{y^3}{6} + \frac{y^5}{120} - \ldots \). For small \( y \), the dominant term is \( y \).

Step 3: Simplify the expression using the leading term of the series

Substitute \( \sin(y) = y - \frac{y^3}{6} + \ldots \) into \( \frac{1}{y} \sin(y) \) to obtain \( \frac{1}{y}(y - \frac{y^3}{6} + \frac{y^5}{120} - \ldots) = 1 - \frac{y^2}{6} + \ldots \).

Step 4: Evaluate the limit as y approaches 0

As \( y \to 0 \), higher order terms like \( -\frac{y^2}{6} \) approach 0 more quickly than the constant term. Thus, the expression simplifies to 1. Therefore, \( \lim_{y \to 0} \left( 1 - \frac{y^2}{6} + \ldots \right) = 1 \).

Key Concepts

Taylor series expansion

The Taylor series is a powerful tool in calculus that allows us to approximate complex functions through polynomials. It provides us with an infinite sum which represents functions in terms of their derivatives at a single point. This is especially useful for functions that are difficult to compute directly, like the sine function when the angle is very small.

The Taylor series expansion of the sine function around 0, or at the point where the angle is zero, is given by:

• \( \sin(y) = y - \frac{y^3}{6} + \frac{y^5}{120} - \dots \)

For small values of \( y \), higher order terms like \( \frac{y^3}{6} \) and \( \frac{y^5}{120} \) become negligible very quickly. Thus, the series can often be truncated after the linear term.

Taylor series are not only useful for approximating sine but also other functions such as exponential and logarithmic functions. This series can give us good approximations when dealing with limits, like \( \lim_{y \to 0} \) where small values of \( y \) make the series terms beyond the first one very small.

Substitution method in calculus

The substitution method is a common technique in calculus for simplifying expressions, especially when evaluating limits. By defining a new variable that replaces part of the original expression, we can reduce a complex problem into a more manageable one.

In the original exercise, we substitute \( y = \frac{1}{x+1} \) which simplifies the problem significantly. As \( x \to \infty \), \( y \to 0^+ \). This substitution transforms the original expression \((x+1) \sin \frac{1}{x+1}\) into \( \frac{1}{y} \sin(y) \).

• This change of variables turns the potentially challenging limit as \( x \) approaches infinity into a simpler limit as \( y \) approaches zero.
• It often helps to identify hidden limits and reveals the underlying behavior of the function in its neighborhood.
  

By using the substitution method, expressions can be re-written in ways that leverage mathematical tools like the Taylor series to simplify the evaluation.

Sine function

The sine function is one of the basic trigonometric functions that maps an angle to the ratio of the lengths of the sides of a right triangle opposite and hypotenuse. In calculus, the sine function is important for studying periodic behavior and oscillations.

Near zero, the sine function behaves particularly nicely and can be approximated as its angle itself, that is, \( \sin(y) \approx y \). This is because the sine of small angles is nearly equal to the line y = x. This property is exploited through the use of its Taylor series expansion.

• When thinking of limits like \( \lim_{x \to 0} \sin(x)/x \), the small angle approximation helps: it equals 1.
• This approximation greatly simplifies the evaluation of expressions where \( y \), a small angle, is involved.

Understanding the properties of the sine function helps in solving problems related to wave mechanics, sound waves, and other applications involving periodic motion. In our exercise, using these properties shows how a complex expression can reduce to a simpler form.