Question

1. for the function g graphed in the accompanying figure, find (a) $limlimits_{x \to 4^-} g(x)$ (b) $limlimits_{x \to 4^+} g(x)$ (c) $limlimits_{x \to 4} g(x)$ (d) $g(4)$ (e) $limlimits_{x \to -\infty} g(x)$ (f) $limlimits_{x \to +\infty} g(x)$. figure ex-3

Explanation:

Part (a): $\boldsymbol{\lim_{x \to 4^-} g(x)}$

Step1: Analyze left - hand limit

To find the left - hand limit as $x$ approaches $4$ (i.e., $x\to4^-$), we look at the values of $g(x)$ as $x$ gets closer to $4$ from the left side (values less than $4$). From the graph, as $x$ approaches $4$ from the left, the function $g(x)$ approaches the value $1$.

Step1: Analyze right - hand limit

To find the right - hand limit as $x$ approaches $4$ (i.e., $x\to4^+$), we look at the values of $g(x)$ as $x$ gets closer to $4$ from the right side (values greater than $4$). From the graph, as $x$ approaches $4$ from the right, the function $g(x)$ approaches the value $1$.

Step1: Recall the limit existence condition

The limit $\lim_{x \to a}f(x)$ exists if and only if $\lim_{x \to a^-}f(x)=\lim_{x \to a^+}f(x)$.

Step2: Use results from (a) and (b)

We know from part (a) that $\lim_{x \to 4^-} g(x) = 1$ and from part (b) that $\lim_{x \to 4^+} g(x)=1$. Since the left - hand limit and the right - hand limit are equal, $\lim_{x \to 4} g(x)$ exists and is equal to $1$.

Answer:

$\lim_{x \to 4^-} g(x)=1$

Part (b): $\boldsymbol{\lim_{x \to 4^+} g(x)}$