Question

In Exercises \(25-34,\) find the limit. $$ \lim _{x \rightarrow 0^{+}} e^{-0.5 x} \sin x $$

Short answer

The limit as \(x\) approaches \(0^{+}\) of \(e^{-0.5x} \sin x\) is \(0\).

Step-by-step solution

Identify Indeterminate Form

First, to identify the form as \(0 \times \infty\) we substitute \(x\) as \(0\) into the equation, yielding \(e^{-0.5(0)} \cdot \sin(0) = 1 \cdot 0 = 0\). The result is an indeterminate form. As such, we will need additional steps to calculate this limit.

Transforming Indeterminate form to \(0/0\) or \(\infty/\infty\)

The goal is to get a form where L'Hopital's Rule can be applied. L'Hopital's Rule is used to evaluate limits of the form \(0/0\) or \(\infty/\infty\). Rewrite the expression as \(\frac{e^{-0.5x}}{1/\sin(x)}\). Then, substituting in \(x = 0^{+}\) results in \(0/0\) – a form in which we can apply L'Hopital’s rule.

Applying L'Hopital’s rule

By L'Hopital’s rule, \(\lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}\), where \(f(x) = e^{-0.5x}\) and \(g(x) = 1/\sin(x)\). The derivatives of \(f(x)\) and \(g(x)\) are \(-0.5e^{-0.5x}\) and \(-\csc^2(x)\) respectively. Thus, the new limit as \(x\) approaches \(0^{+}\) becomes \(\frac{-0.5e^{-0.5x}}{-\csc^2(x)} = 0.5e^{-0.5x} \sin^2(x)\).

Evaluate the new limit

Now, we substitute \(x = 0\) into the new expression to determine the limit value, resulting in \(0.5e^{-0.5(0)} \sin^2(0) = 0.5(1) * 0 = 0\).