Question

Find the following limits using Maclaurin series and check your results by computer. Hint: First combine the fractions. Then find the first term of the denominator series and the first term of the numerator series. $$\text { (a) } \lim _{x \rightarrow 0}\left(\frac{1}{x}-\frac{1}{e^{x}-1}\right)$$ $$\text { (b) } \lim _{x \rightarrow 0}\left(\frac{1}{x^{2}}-\frac{\cos x}{\sin ^{2} x}\right)$$ $$\text { (c) } \lim _{x \rightarrow 0}\left(\csc ^{2} x-\frac{1}{x^{2}}\right)$$ $$\text { (d) } \lim _{x \rightarrow 0}\left(\frac{\ln (1+x)}{x^{2}}-\frac{1}{x}\right)$$

Short answer

a) \(\frac{1}{2}\) b) 0 c) 0 d) -\(\frac{1}{2}\)

Step-by-step solution

Simplify the expression (a)

Combine the fractions: \[ \lim_{x \rightarrow 0} \left( \frac{1}{x} - \frac{1}{e^x - 1} \right) = \lim_{x \rightarrow 0} \left( \frac{e^x - 1 - x}{x(e^x - 1)} \right) \]

Maclaurin series for numerator (a)

The Maclaurin series for \(e^x\) is \(1 + x + \frac{x^2}{2!} + \cdots\), so we get: \( e^x - 1 - x \approx \frac{x^2}{2!} = \frac{x^2}{2}\)

Maclaurin series for denominator (a)

The denominator is \(x(e^x - 1)\). Using the Maclaurin series for \(e^x - 1\), we get: \( x(e^x - 1) \approx x(x + \frac{x^2}{2!}) = x^2 + \frac{x^3}{2!} \approx x^2 \)

Compute limit (a)

Using the simplified series: \( \lim_{x \rightarrow 0} \frac{\frac{x^2}{2}}{x^2} = \lim_{x \rightarrow 0} \frac{1}{2} = \frac{1}{2} \)

Simplify the expression (b)

Combine the fractions: \( \lim_{x \rightarrow 0} \left( \frac{1}{x^2} - \frac{\cos x}{\sin^2 x} \right) \)

Maclaurin series for numerator (b)

For small \(x\), \( \cos x \approx 1 \). So we get: \( \frac{1}{x^2} - \frac{1}{\sin^2 x} \approx \frac{1}{x^2} - \frac{1}{x^2} = 0 \)

Simplify the expression (c)

The cosecant function is \( \csc x = \frac{1}{\sin x} \). So we need to find: \( \lim_{x \rightarrow 0} \left( \csc^2 x - \frac{1}{x^2} \right) \)

Maclaurin series for numerator (c)

For small \(x\), \( \sin x \approx x \). So we get: \( \csc^2 x = \left( \frac{1}{\sin x} \right)^2 \approx \left( \frac{1}{x} \right)^2 = \frac{1}{x^2} \). Thus, \( \csc^2 x - \frac{1}{x^2} = 0 \)

Simplify the expression (d)

Combine the fractions: \( \lim_{x \rightarrow 0} \left( \frac{\ln(1 + x)}{x^2} - \frac{1}{x} \right) = \lim_{x \rightarrow 0} \left( \frac{\ln(1 + x) - x}{x^2} \right) \)

Maclaurin series for numerator (d)

Using the Maclaurin series for \(\ln(1 + x)\), we get: \( \ln(1 + x) \approx x - \frac{x^2}{2} \). So: \( \ln(1 + x) - x \approx -\frac{x^2}{2} \)

Compute limit (d)

Using the simplified series: \( \lim_{x \rightarrow 0} \frac{-\frac{x^2}{2}}{x^2} = \lim_{x \rightarrow 0} -\frac{1}{2} = -\frac{1}{2} \)

Key concepts

Maclaurin series

The Maclaurin series is essentially a Taylor series expansion of a function about zero. This allows us to express functions as infinite sums of terms. Each term is derived from the function's derivatives at zero. For instance, the Maclaurin series for the exponential function, denoted as \( e^x \), is: \[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \] This expansion is very useful for approximating functions when \( x \) is close to zero. In our limit calculations, we use these series to simplify complicated expressions. By taking just the first few terms, we approximate the function's behavior near zero with great accuracy. Always remember that these approximations are most effective when \( x \) is very small.

Limit calculation

Calculating limits involves finding the value that a function approaches as the input approaches a certain value. In the context of Maclaurin series, we often use these expansions to simplify difficult functions and make the limit calculations more manageable. For example, consider the limit: \[ \lim_{x \rightarrow 0} \left( \frac{1}{x} - \frac{1}{e^x - 1} \right) \] First, we combine the fractions: \[ \lim_{x \rightarrow 0} \left( \frac{e^x - 1 - x}{x(e^x - 1)} \right) \] By using the Maclaurin series for \( e^x \), we can simplify \( e^x - 1 - x \) to \( \frac{x^2}{2} \). Similarly, the denominator \( x(e^x - 1) \) simplifies to \( x^2 \). Thus, the limit becomes: \[ \lim_{x \rightarrow 0} \frac{\frac{x^2}{2}}{x^2} = \frac{1}{2} \] In essence, breaking down the function using series expansion simplifies the process and makes the evaluation of the limit clearer and more intuitive.

Calculus

Calculus, the mathematical study of continuous change, is foundational for understanding complex behaviors in functions. Two central concepts in calculus are differentiation and integration. Differentiation looks at the instantaneous rate of change of a function, while integration sums up all parts to understand the whole. The exercises provided involve differentiation, particularly how functions change as x approaches zero. Calculus provides a systematic way to calculate these changes. For example, when finding: \[ \lim_{\ x \rightarrow 0} \left( \frac{\ln(1 + x)}{x^2} - \frac{1}{x} \right) \] First, it's simplified using the Maclaurin series: \[ \lim_{x \rightarrow 0} \left( \frac{\ln(1 + x) - x}{x^2} \right) \] By substituting the series expansion for \( \ln(1 + x) \) and simplifying, we get: \[ \ln(1 + x) \approx x - \frac{x^2}{2} \] and thus: \[ \ln(1 + x) - x \approx -\frac{x^2}{2} \] Finally, simplifying the limit yields: \[ \lim_{x \rightarrow 0} -\frac{1}{2} = -\frac{1}{2} \] Mastery of these techniques and understanding their applications are indispensable tools in calculus. They are widely used in fields like physics, engineering, economics, and beyond.