Question

Use the graph of the function \(f\) to state the value of each limit, if it exists. If it does not exist, explain why.

(a)\(\mathop {{\bf{lim}}}\limits_{{\bf{x}} \to {{\bf{0}}^ - }} f\left( x \right)\) (b)\(\mathop {{\bf{lim}}}\limits_{{\bf{x}} \to {{\bf{0}}^{\bf{ + }}}} f\left( x \right)\) (c)\(\mathop {{\bf{lim}}}\limits_{{\bf{x}} \to {\bf{0}}} f\left( x \right)\)

14. \(f\left( x \right){\bf{ = }}\frac{{{e^{{\bf{1}}/x}} - {\bf{2}}}}{{{e^{{\bf{1}}/x}} + {\bf{1}}}}\)

Short answer

1. \(\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = - 2\)
2. \(\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = 1\)
3. \(\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = {\rm{does}}\;{\rm{not}}\,{\rm{exist}}\)

Step-by-step solution

Plot the graph of the function

Draw the graph of the function \(f\left( x \right) = \frac{{{e^{1/x}} - 2}}{{{e^{1/x}} + 1}}\) by using the graphing calculator as:

1. Open the graphing calculator. Select the “STAT PLOT” and enter the equation \(\left( {{e^{1/X}} - 2} \right)/\left( {{e^{1/X}} + 1} \right)\) in the \({Y_1}\) tab.
2. Enter the “GRAPH” button in the graphing calculator.

The graph of the function \(f\left( x \right) = \frac{{{e^{1/x}} - 2}}{{{e^{1/x}} + 1}}\) is shown below:

a) Step 2: Observe the graph when \(x\) tends to \({{\bf{0}}^ - }\)

From the graph, it is observed that the value of the function tends to \( - 2\) when \(x\) tends to \({0^ - }\).

Therefore, \(\mathop {\lim }\limits_{x \to {0^ - }} \frac{{{e^{1/x}} - 2}}{{{e^{1/x}} + 1}} = - 2\).

b) Step 3: Observe the graph when \(x\) tends to \({{\bf{0}}^{\bf{ + }}}\)

From the graph, it is observed that the value of function tends to \(1\) when \(x\) tends to \({0^ + }\).

Therefore, \(\mathop {\lim }\limits_{x \to {0^ + }} \frac{{{e^{1/x}} - 2}}{{{e^{1/x}} + 1}} = 1\).

c) Step 4: Observe the graph when \({\bf{x}}\) tends to \({\bf{0}}\)

Use the result,\(\mathop {\lim }\limits_{x \to a} f\left( x \right) = L\)if and if only\(\mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right) = L\)and\(\mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right) = L\).

From the parts (a) and (b), we get\(\mathop {\lim }\limits_{x \to {0^ - }} \frac{{{e^{1/x}} - 2}}{{{e^{1/x}} + 1}} = - 2\) and \(\mathop {\lim }\limits_{x \to {0^ + }} \frac{{{e^{1/x}} - 2}}{{{e^{1/x}} + 1}} = 1\).

Since both left- hand limit and right-hand limit are not equal. This implies that limit of function at \(x = 0\) does not exist.

Therefore, \(\mathop {\lim }\limits_{x \to 0} \frac{{{e^{1/x}} - 2}}{{{e^{1/x}} + 1}} = {\rm{does}}\;{\rm{not}}\,{\rm{exist}}\).