Question

In Exercises \(21-26,\) find \(\lim _{x \rightarrow \infty} y\) and \(\lim _{x \rightarrow-\infty} y\). $$y=\frac{x \sin x+2 \sin x}{2 x^{2}}$$

Short answer

The limit as \(x\) approaches \(\infty\) or \(-\infty\) is undefined.

Step-by-step solution

Identify Limit Forms

First identify if the limit form is an indeterminate form of type \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). Substituting \(x=\infty\) and \(x=-\infty\) into the given function \(y=\frac{x \sin x+2 \sin x}{2 x^{2}}\) provides these forms.

Apply L'Hospital's Rule once

Apply L'Hospital's rule by taking the derivative of the numerator and denominator. The derivative of the numerator \(xsinx + 2 sinx\) becomes, using product rule and constant rule, \( xsin(x) \to cosx - xsinx\) and \(2sinx \to 2cosx\). Therefore, the derivative of the complete numerator is \( cosx - xsinx + 2cosx\). The derivative of the denominator \(2x^2\) is \(4x\). The function becomes \(\frac{cosx - xsinx + 2cosx}{4x}\).

Analyze new function

Check if the new function still has the limit form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). If it does, then apply L'Hospital's rule again. If not, then you may calculate the limit normally.

Apply L'Hospital's Rule again

The application of L'Hospital's rule again gives us \(\frac{-sinx - sinx - xcosx + 2(-sinx)}{4}\). This given function simplifies to \(\frac{-3sinx - xcosx}{4}\). Note that as \(x -> \infty\) or \(x -> -\infty\), the sinusoidal terms will oscillate between -1 and 1.

Compute the Limit

Now, compute the limit as \(x \rightarrow \infty\) and \(x\rightarrow-\infty\). As \(x\) approaches either \(\infty\) or \(-\infty\), \(-3sinx - xcosx\) will also oscillate between positive and negative values. The limit is therefore undefined as \(x\) tends to \(\pm\infty\).