Problem 66 - Calculus Early Transcendentals

Question

Use Taylor series to evaluate the following limits. Express the result in terms of the parameter(s). $$\lim _{x \rightarrow 0} \frac{e^{a x}-1}{x}$$

Step-by-Step Solution

Answer

Question: Determine the limit of the given expression involving an exponential function using Taylor series: $$\lim_{x\rightarrow0}\frac{e^{ax} - 1}{x}$$ Answer: The limit of the given expression, expressed in terms of the parameter 'a', is: $$\lim_{x\rightarrow0}\frac{e^{ax} - 1}{x} = a$$

Step 1: Identify the function to expand as Taylor series

The function to expand as Taylor series is: $$f(x) = e^{ax}$$

Step 2: Write down the Taylor series expansion of the function

The Taylor series expansion of the function \(f(x) = e^{ax}\) about the point x=0 is given by: $$e^{ax} = 1 + ax + \frac{1}{2}(ax)^2 + \frac{1}{6}(ax)^3 + \dots$$

Step 3: Calculate the expression in the limit using the Taylor series expansion

Now we have to compute the limit of the given expression using the expanded Taylor series. We will replace the function \(f(x)\) in the expression with its Taylor series expansion: $$\lim _{x \rightarrow 0} \frac{e^{a x}-1}{x} = \lim _{x \rightarrow 0} \frac{1+ax+\frac{1}{2}(ax)^2+\frac{1}{6}(ax)^3+\dots-1}{x}$$ Simplify the expression by canceling out the 1 in the numerator: $$\lim _{x \rightarrow 0} \frac{ax+\frac{1}{2}(ax)^2+\frac{1}{6}(ax)^3+\dots}{x}$$ Divide the whole expression by x: $$\lim _{x \rightarrow 0} a+\frac{1}{2}ax+\frac{1}{6}(ax)^2+\dots$$

Step 4: Evaluate the limit

Now we can evaluate the limit as x approaches 0: $$\lim _{x \rightarrow 0} a+\frac{1}{2}ax+\frac{1}{6}(ax)^2+\dots = a$$ Hence, the final result of the limit, expressed in terms of parameter 'a', is: $$\lim _{x \rightarrow 0} \frac{e^{a x}-1}{x} = a$$