Question

$$lim_{x \to -\frac{pi}{2}} \tan(x) =?$$
choose 1 answer:
a $-1$
b $-\frac{1}{2}$
c $0$
d the limit doesnt exist.

Explanation:

Step1: Recall the definition of tangent function

The tangent function is defined as $\tan(x) = \frac{\sin(x)}{\cos(x)}$. We need to find the limit as $x$ approaches $-\frac{\pi}{2}$.

Step2: Analyze the behavior of $\cos(x)$ and $\sin(x)$ as $x \to -\frac{\pi}{2}$

• For $\cos(x)$: As $x \to -\frac{\pi}{2}$, $\cos(x) \to 0$. Specifically, when approaching from the left side of $-\frac{\pi}{2}$ (i.e., $x \to -\frac{\pi}{2}^-$), $\cos(x)$ approaches $0$ from the positive side (since $\cos(-\frac{\pi}{2}+\epsilon)=\sin(\epsilon)\approx\epsilon>0$ for small $\epsilon>0$). When approaching from the right side of $-\frac{\pi}{2}$ (i.e., $x \to -\frac{\pi}{2}^+$), $\cos(x)$ approaches $0$ from the negative side (since $\cos(-\frac{\pi}{2}-\epsilon)=-\sin(\epsilon)\approx-\epsilon<0$ for small $\epsilon>0$).
• For $\sin(x)$: As $x \to -\frac{\pi}{2}$, $\sin(x) \to - 1$.

Step3: Analyze the left - hand limit and right - hand limit of $\tan(x)$

• Left - hand limit ($x \to -\frac{\pi}{2}^-$): $\tan(x)=\frac{\sin(x)}{\cos(x)}$, $\sin(x)\to - 1$ and $\cos(x)\to0^+$. So, $\lim_{x\to-\frac{\pi}{2}^-}\tan(x)=\frac{-1}{0^+}=-\infty$.
• Right - hand limit ($x \to -\frac{\pi}{2}^+$): $\tan(x)=\frac{\sin(x)}{\cos(x)}$, $\sin(x)\to - 1$ and $\cos(x)\to0^-$. So, $\lim_{x\to-\frac{\pi}{2}^+}\tan(x)=\frac{-1}{0^+}=\infty$ (because $\frac{-1}{0^-}=\frac{-1}{-|0|}=\frac{-1}{-0^+}=\infty$).

Since the left - hand limit ($-\infty$) and the right - hand limit ($\infty$) are not equal, the two - sided limit $\lim_{x\to-\frac{\pi}{2}}\tan(x)$ does not exist.

Answer:

D. The limit doesn't exist.