Solveeit Logo

Question

Question: In the following graph how do you determine the value of c such that \[\displaystyle \lim_{x \to c}f...

In the following graph how do you determine the value of c such that limxcf(x)\displaystyle \lim_{x \to c}f\left( x \right) exists?

Explanation

Solution

Here we have to determine the value of c such that limxcf(x)\displaystyle \lim_{x \to c}f\left( x \right) exists. We can see that in the graph the function f(x)f\left( x \right) is defined when x = -2 but the value which f(x)f\left( x \right) will approach as x gets closer to -3 from the left is different from the value as it approaches from the right. We can apply the limits by looking at the graph and find the c values.

Complete step by step solution:
Here we have to determine the value of c such that limxcf(x)\displaystyle \lim_{x \to c}f\left( x \right) exists.
We can see that in the graph the function f(x)f\left( x \right) is defined when x = -2 but the value which f(x)f\left( x \right) will approach as x gets closer to -3 from the left is different from the value as it approaches from the right.

We can now analyse the graph, as x approaches -3 from left f(x)f\left( x \right) approaches (negative 2) however as x approaches -3 from the right f(x)f\left( x \right) approaches (negative 3)
We can now apply the limits, we get

& \Rightarrow \displaystyle \lim_{x \to -{{3}^{+}}}=-3 \\\ & \Rightarrow \displaystyle \lim_{x \to -{{3}^{-}}}=-2 \\\ \end{aligned}$$ Where the limit does not exist at x = -3. We can now see that, in the same way when x tends to 0, we get $$\begin{aligned} & \Rightarrow \displaystyle \lim_{x \to {{0}^{+}}}=1 \\\ & \Rightarrow \displaystyle \lim_{x \to {{0}^{-}}}=+\infty \\\ \end{aligned}$$ **So, clearly we can assume that c = -3 or c = 0, but the limit does not exist.** **Note:** here we have already given a graph, from which we can analyse the numbers to be found and we can assume the limit values to find the value for what we have been asked for. We should also know how to apply the limit concept to solve these types.