Question

Explain what it means to say that

\(\mathop {{\bf{lim}}}\limits_{x \to {{\bf{1}}^ - }} f\left( x \right) = {\bf{3}}\)and \(\mathop {{\bf{lim}}}\limits_{x \to {{\bf{1}}^ + }} f\left( x \right) = {\bf{7}}\)

In this situation, is it possible that\(\mathop {{\bf{lim}}}\limits_{x \to {\bf{1}}} f\left( x \right)\) exists? Explain.

Short answer

The expressions \(\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = 3\) and \(\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = 7\) represent that value of f is 3 and 7 when x is approaching 1 from left and right, respectively.

Step-by-step solution

Step 1:Interpretation of the equation

The equation\(\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = 3\) represents that as x is approaching 1 from the left, the value of \(f\left( x \right)\) will approach3.

The equation \(\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = 7\) represents that as x approachesone from right, the value of \(f\left( x \right)\) will approach 7.

Check whether \(\mathop {{\bf{lim}}}\limits_{x \to {\bf{1}}} f\left( x \right)\) exists

As at \(x = 1\) the left hand and right-hand limitsof \(f\left( x \right)\)are different, therefore the limit at \(x = 1\) does not exist for \(f\left( x \right)\).