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), $f(x)=e^{x}\tan(\frac{3x}{4})$.

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Step2: Find right - hand limit

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

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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 $x=\pi$.