Question

use the graph of the function f to state the value of each limit, if it exists. (if an answer does not exist, enter dne.)

f(x) = \frac{e^{1/x} - 4}{e^{1/x} + 1}

(a) \lim\limits_{x \to 0^-} f(x)
\boxed{1.5} \times

(b) \lim\limits_{x \to 0^+} f(x)
\boxed{1.5} \times
enter an exact number.

(c) \lim\limits_{x \to 0} f(x)
\boxed{-1.5} \times

Explanation:

Part (a): $\boldsymbol{\lim_{x \to 0^-} f(x)}$

Step 1: Analyze $1/x$ as $x \to 0^-$

As $x$ approaches $0$ from the left ($x \to 0^-$), $x$ is negative and approaches $0$, so $\frac{1}{x} \to -\infty$.

Step 2: Evaluate $e^{1/x}$ as $1/x \to -\infty$

Recall that $\lim_{t \to -\infty} e^t = 0$. Let $t = \frac{1}{x}$, so as $x \to 0^-$, $t \to -\infty$, and $e^{1/x} \to 0$.

Step 3: Substitute into $f(x)$

Substitute $e^{1/x} \to 0$ into $f(x) = \frac{e^{1/x} - 4}{e^{1/x} + 1}$:

$$ \lim_{x \to 0^-} \frac{e^{1/x} - 4}{e^{1/x} + 1} = \frac{0 - 4}{0 + 1} = -4 $$

Part (b): $\boldsymbol{\lim_{x \to 0^+} f(x)}$

Step 1: Analyze $1/x$ as $x \to 0^+$

As $x$ approaches $0$ from the right ($x \to 0^+$), $x$ is positive and approaches $0$, so $\frac{1}{x} \to +\infty$.

Step 2: Evaluate $e^{1/x}$ as $1/x \to +\infty$

Recall that $\lim_{t \to +\infty} e^t = +\infty$. Let $t = \frac{1}{x}$, so as $x \to 0^+$, $t \to +\infty$, and $e^{1/x} \to +\infty$.

Step 3: Simplify $f(x)$ for large $e^{1/x}$

For large $e^{1/x}$, we can divide numerator and denominator by $e^{1/x}$:

$$ f(x) = \frac{e^{1/x} - 4}{e^{1/x} + 1} = \frac{1 - \frac{4}{e^{1/x}}}{1 + \frac{1}{e^{1/x}}} $$

As $e^{1/x} \to +\infty$, $\frac{4}{e^{1/x}} \to 0$ and $\frac{1}{e^{1/x}} \to 0$. Thus:

$$ \lim_{x \to 0^+} \frac{1 - \frac{4}{e^{1/x}}}{1 + \frac{1}{e^{1/x}}} = \frac{1 - 0}{1 + 0} = 1 $$

Part (c): $\boldsymbol{\lim_{x \to 0} f(x)}$

Step 1: Check left and right limits

From part (a), $\lim_{x \to 0^-} f(x) = -4$, and from part (b), $\lim_{x \to 0^+} f(x) = 1$.

Step 2: Determine if the limit exists

For the two - sided limit $\lim_{x \to 0} f(x)$ to exist, the left - hand limit and the right - hand limit must be equal. Since $-4
eq1$, $\lim_{x \to 0} f(x)$ does not exist (DNE).

Answer:

s:
(a) $\boldsymbol{-4}$
(b) $\boldsymbol{1}$
(c) $\boldsymbol{DNE}$