evaluate the following: $lim_{x\rightarrow0}\frac{cos(x)}{x - cos(x)}$\n-∞\n0\ndne\n1\n-1\nquestion 3\n1…

Question 1

Explanation:

Step1: Substitute $x = 0$

Substitute $x = 0$ into $\frac{\cos(x)}{x-\cos(x)}$. We get $\frac{\cos(0)}{0 - \cos(0)}$. Since $\cos(0)=1$, the expression becomes $\frac{1}{0 - 1}=-1$.

Answer:

-1

Question 2

Explanation:

Step1: Analyze left - hand and right - hand limits

For $\lim_{x\rightarrow2}\frac{3}{x - 2}$, as $x\rightarrow2^{+}$ (approaching 2 from the right), $x-2>0$ and $x - 2\rightarrow0$. So $\frac{3}{x - 2}\rightarrow+\infty$. As $x\rightarrow2^{-}$ (approaching 2 from the left), $x - 2<0$ and $x - 2\rightarrow0$. So $\frac{3}{x - 2}\rightarrow-\infty$. Since the left - hand limit and the right - hand limit are not equal, the limit does not exist (DNE).

Answer:

DNE