Question

In Exercises 9-12, find the limit and confirm your answer using the Sandwich Theorem. $$\lim _{x \rightarrow-\infty} \frac{1-\cos x}{x^{2}}$$

Short answer

The limit of the function \(\frac{1-\cos x}{x^{2}}\) as \(x\) approaches negative infinity is \(0\).

Step-by-step solution

Finding Limit

Firstly, the limit of the given function should be found as \(x\) approaches negative infinity. Due to the complexity of the function, it should be split into two parts to ease calculation:\[\lim_{x \rightarrow-\infty} \frac{1}{x^{2}}- \lim_{x \rightarrow-\infty}\frac{\cos x}{x^{2}}.\]The limit of \(\frac{1}{x^{2}}\) as \(x\) approaches negative infinity is zero. The value of cosine doesn't depend on \(x\) and since \(x\) approaches negative infinity, \(\frac{\cos x}{x^{2}}\) will also tend to zero. This gives limit as \(0-0=0\).

Confirming using Sandwich Theorem

The Sandwich Theorem can now be utilized to verify the obtained answer. It should be noted that \(-1 \leq \cos x \leq 1\). As such, for the negative fraction, the lower limit will be \(\frac{1-1}{x^2}=0\) and the upper limit will be \(\frac{1+1}{x^2}= \frac{2}{x^{2}}\).As \(x\) approaches negative infinity, both the lower and the upper boundaries will tend to zero as well, confirming that the original function must approach zero too, as it is 'sandwiched' between these two values.