Question

evaluate the following: $lim_{x
ightarrow0}\frac{cos(x)}{x - cos(x)}$
-∞
0
dne
1
-1
question 3
1 pts
evaluate the following: $lim_{x
ightarrow2}\frac{3}{x - 2}$
0
∞
-∞
1
dne

Explanation:

Question 1

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 $\frac{1}{0 - 1}=-1$.

Step1: Analyze left - hand and right - hand limits

For $\lim_{x
ightarrow2}\frac{3}{x - 2}$, as $x
ightarrow2^{+}$ (approaching 2 from the right), $x-2>0$ and $x - 2
ightarrow0$. So $\frac{3}{x - 2}
ightarrow+\infty$.
As $x
ightarrow2^{-}$ (approaching 2 from the left), $x - 2<0$ and $x - 2
ightarrow0$. So $\frac{3}{x - 2}
ightarrow-\infty$.
Since the left - hand limit and the right - hand limit are not equal, the limit does not exist (DNE).

Answer:

-1

Question 2