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) \(\mathop {{\bf{lim}}}\limits_{{\bf{x}} \to {\bf{0}}} f\left( x \right)\)

13. \(f\left( x \right){\bf{ = }}x\sqrt {{\bf{1}} + {x^{ - {\bf{2}}}}} \)

Short answer

(a) \(\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = - 1\)

(b) \(\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = 1\)

(c) \(\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = {\rm{does}}\;{\rm{not}}\,{\rm{exist}}\)

Step-by-step solution

Plot the graph of the function

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

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

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

(a)

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

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

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

(b)

Observe the graph when \({\bf{x}}\) 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, \(\mathop {\lim }\limits_{x \to {0^ + }} x\sqrt {1 + {x^{ - 2}}} = 1\).

(c)

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 part (a) and (b), we get\(\mathop {\lim }\limits_{x \to {0^ - }} x\sqrt {1 + {x^{ - 2}}} = - 1\) and \(\mathop {\lim }\limits_{x \to {0^ + }} x\sqrt {1 + {x^{ - 2}}} = 1\).

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

Therefore, \(\mathop {\lim }\limits_{x \to 0} f\left( x \right) = {\rm{does}}\;{\rm{not}}\,{\rm{exist}}\).