Question

Sketch the graph of the following function and use it to determine the values of a for which\(\mathop {lim}\limits_{x \to a} f\left( x \right)\)exists:

\(f\left( x \right) = \left\{ \begin{array}{l}1 + sinx if x < 0\\cosx if 0 \le x \le \pi \\sinx ifx > \pi \end{array} \right\}\)

Short answer

For the data, \(f\left( x \right) = \left\{ \begin{array}{l}1 + sinx if x < 0\\cosx if 0 \le x \le \pi \\sinx ifx > \pi \end{array} \right\}\) the graph is presented below.

\(\mathop {\lim }\limits_{x \to a} f\left( x \right)\)exist for all values of a which lies in the interval, \(\left( { - \infty ,\pi } \right) \cup \left( {\pi ,\infty } \right)\).

Step-by-step solution

Introduction

In one sided limit,

\(\mathop {\lim }\limits_{x \to {0^ - }} H(t) = 0{\rm{ }}and{\rm{ }}\mathop {\lim }\limits_{x \to {0^ + }} H(t) = 1\)

The symbol\(x \to {0^ - }{\rm{ and }}x \to {0^ + }{\rm{ }}\)indicates that considered values are less than zero and greater than zero respectively.

The negative sign (-) value will be gotten from the left side and the positive sign value (+) will be gotten from the right side.

If both the values will be equal then the value of \(x \to 0\)will be also equal to both the values and if both the values are unequal then \(x \to 0\)won’t be equal to any of them.

Given information

The given function is,\(f\left( x \right) = \left\{ \begin{array}{l}1 + sinx if x < 0\\cosx if 0 \le x \le \pi \\sinx ifx > \pi \end{array} \right\}\).

Graph

For the given data the graph is as following, the \(\sin x\) graph starts after point 3 because its value of it is more than 180 degrees. The \(\cos x\) will start from zero according to their limit and \(1 + \sin x\)will start from the negative side as it is lower than zero.

Where,\(f\left( x \right) = \left\{ \begin{array}{l}Red line is , 1 + sinx if x < 0\\Blue line is, cosx if 0 \le x \le \pi \\Green line is, sinx ifx > \pi \end{array} \right\}\).

From the graph, it can be observed that,\(\mathop {\lim }\limits_{x \to a} f\left( x \right)\)does not exists for \(x = \pi \).

So, \(\mathop {\lim }\limits_{x \to a} f\left( x \right)\) exist for all values of a which lies in the interval, \(\left( { - \infty ,\pi } \right) \cup \left( {\pi ,\infty } \right)\).