4) $lim_{x \to -infty}-3xcos\frac{1}{x}$

Explanation:

Step1: Analyze the term $\cos\left(\frac{1}{x}\right)$ as $x \to -\infty$.

Let $u = \frac{1}{x}$. As $x \to -\infty$, $u \to 0$. Then, the limit of $\cos\left(\frac{1}{x}\right)$ becomes: $$ \lim_{x \to -\infty} \cos\left(\frac{1}{x}\right) = \lim_{u \to 0} \cos(u) = \cos(0) = 1 $$

Step2: Analyze the term $-3x$ as $x \to -\infty$.

The limit of $-3x$ as $x \to -\infty$ is: $$ \lim_{x \to -\infty} -3x = -3(-\infty) = +\infty $$

Step3: Combine the results to find the overall limit.

We need to evaluate the limit of the product: $$ \lim_{x \to -\infty} -3x \cos\left(\frac{1}{x}\right) $$ Using the results from Step 1 and Step 2: $$ \left(\lim_{x \to -\infty} -3x\right) \cdot \left(\lim_{x \to -\infty} \cos\left(\frac{1}{x}\right)\right) = (+\infty) \cdot 1 $$ $$ = +\infty $$

Answer:

$$ +\infty $$