Question

For the function gwhose graph is given, state the value of each quantity, if it exists. If it does not exist, explain why.

\(\begin{array}{l}(a)\mathop {lim}\limits_{t \to {0^ - }} g(t) (b)\mathop {lim}\limits_{t \to {0^ + }} g(t) (c) \mathop {lim}\limits_{t \to 0} g(t)\\(d)\mathop {lim}\limits_{t \to {2^ - }} g(t) (e)\mathop {lim}\limits_{t \to {2^ + }} g(t) (f) \mathop {lim}\limits_{t \to 2} g(t)\\(g) g(2) (h)\mathop {lim}\limits_{t \to 4} g(t) \end{array}\)

Short answer

to {2^ + }} g(t) = 0\)

(f)The value of \(\mathop {\lim }\limits_{t \to 2} g(t)\)does not exist.

(g) \(g\left( 2 \right) = 1\)

(h) The value of \(\mathop {\lim }\limits_{t \to 4} g(t)\)is 3.

Step-by-step solution

Limit of a function

The limit of a function at a point exists if the value of the left-hand limit and right-hand limit is equal.

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

Step 2: \((a)\,\,\mathop {lim}\limits_{t \to {0^ - }} g(t) \)

From the graph, it can be seen that when\(t\)to tends to 0 from the left side then the value of \(y\)approaches to -1.

Therefore,

\(\mathop {\lim }\limits_{t \to {0^ - }} g(t) = - 1\)

Step 3: \((b)\mathop {lim}\limits_{t \to {0^ + }} g(t) \)

From the graph, it can be seen that when\(t\)to tends to 0 from the right side then the value of \(y\)approaches to -2.

Therefore,

\(\mathop {\lim }\limits_{t \to {0^ - }} g(t) = - 2\)

\(\)

Step 4: \((c) \mathop {lim}\limits_{t \to 0} g(t)\)

The value of\(\mathop {\lim }\limits_{t \to {0^ + }} g(t)\) and \(\mathop {\lim }\limits_{t \to {0^ - }} g(t)\) is not equal, hence the value of \(\mathop {\lim }\limits_{t \to 0} g(t)\)does not exist.

Step 5: \((d)\mathop {lim}\limits_{t \to {2^ - }} g(t) \)

From the graph, it can be seen that when\(t\)to tends 2 from the left side then the value of \(y\)approaches to 2.

Therefore,

\(\mathop {\lim }\limits_{t \to {2^ - }} g(t) = 2\)

Step 6: \((e)\mathop {lim}\limits_{t \to {2^ + }} g(t) \)

From the graph, it can be seen that when\(t\)to tends 2 from the right side then the value of \(y\)approaches to 0.

Therefore,

\(\mathop {\lim }\limits_{t \to {2^ + }} g(t) = 0\)

Step 7: \((f) \mathop {lim}\limits_{t \to 2} g(t)\)

The value of\(\mathop {\lim }\limits_{t \to {2^ - }} g(t)\) and \(\mathop {\lim }\limits_{t \to {2^ + }} g(t)\) is not equal, hence the value of \(\mathop {\lim }\limits_{t \to 2} g(t)\)does not exist.

\(\)

Step 8:\((g) \,\,\,\,g(2)   \)

From the graph, it can be seen that if the value of\(t\)is 2 then the value of \(y\) is 1.

Therefore,

\(g\left( 2 \right) = 1\)

Step 9:\((h)\mathop {lim}\limits_{t \to 4} g(t)  \)

The value of\(\mathop {\lim }\limits_{t \to {4^ - }} g(t)\) and \(\mathop {\lim }\limits_{t \to {4^ + }} g(t)\)is 3. Since both the values are equal, hence the value of \(\mathop {\lim }\limits_{t \to 4} g(t)\)is 3.