Question

1. $f(x)=\begin{cases}e^{x}cos x, & xgeqpi\\e^{x}\tanleft(\frac{3x}{4}

ight), & x<piend{cases}$ at $x = pi$

Explanation:

Step1: Find left - hand limit

We need to find $\lim_{x
ightarrow\pi^{-}}f(x)$. Since $x
ightarrow\pi^{-}$ (approaching $\pi$ from the left), $f(x)=e^{x}\tan(\frac{3x}{4})$.
\[

$$\begin{align*} \lim_{x ightarrow\pi^{-}}e^{x}\tan(\frac{3x}{4})&=e^{\pi}\tan(\frac{3\pi}{4})\\ &=e^{\pi}\times(- 1)\\ &=-e^{\pi} \end{align*}$$

\]

Step2: Find right - hand limit

We need to find $\lim_{x
ightarrow\pi^{+}}f(x)$. Since $x
ightarrow\pi^{+}$ (approaching $\pi$ from the right), $f(x)=e^{x}\cos x$.
\[

$$\begin{align*} \lim_{x ightarrow\pi^{+}}e^{x}\cos x&=e^{\pi}\cos\pi\\ &=e^{\pi}\times(-1)\\ &=-e^{\pi} \end{align*}$$

\]

Step3: Find function value at $x = \pi$

When $x=\pi$, we use the part of the function for $x\geq\pi$, so $f(\pi)=e^{\pi}\cos\pi=-e^{\pi}$.

Since $\lim_{x
ightarrow\pi^{-}}f(x)=\lim_{x
ightarrow\pi^{+}}f(x)=f(\pi)=-e^{\pi}$, the function $f(x)$ is continuous at $x = \pi$.

Answer:

The function $f(x)$ is continuous at $x=\pi$.