Question

Think About It In Exercises 55-58, L'Hopital's Rule is used incorrectly. Describe the error.\(\begin{aligned} \lim _{x \rightarrow \infty} \operatorname{xec} \operatorname{sen} \frac{1}{x} &=\lim _{x \rightarrow \infty} \frac{\cos (1 / x)}{1 / x} \\ &=\lim _{x \rightarrow \infty} \frac{-\sin (1 / x)]\left(1 / x^{2}\right)}{-1 \times x^{2}} \\ &=0 \end{aligned}\)

Short answer

The error originates from the incorrect application of L'Hopital's Rule where it doesn't apply and an incorrect differential calculation. The corrected answer is the limit does not exist.

Step-by-step solution

Identity the mistake

Verify if the limit problem is in an indeterminate form before applying L'Hopital's Rule. In the Exercise the function is not in the form of 0/0 or ∞/∞, therefore applying L'Hopital's Rule is inappropriate. Additionally, there's an incorrect transformation in the calculations from \( \cos (1 / x) / (1/x) \) to \( -\sin (1 / x) / x^{2} \)

Correct the calculations

The correct derivative of \( \cos(1 / x) \) with respect to \( x \) is \( -\sin(1 / x) / x^2 \) and for \( 1 / x \) is \( -1 / x^2 \), so the transformed equation should be \( \lim _{x \rightarrow \infty} \frac{-\sin (1 / x)/( x^{2})}{-1 / x^{2}} \) which simplifies to \( \lim _{x \rightarrow \infty} \sin(1 / x) \) . This limit does not exist because as \( x \) approaches infinity, \( sin(1 / x) \) will oscilate between -1 and 1.

Correct the answer

Given that the limit doesn't exist, you conclude that the incorrect answer of 0 from the exercise is due to the wrong application of L'Hopital's Rule