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Section 2-5 : Computing Limits

10. Given the function

\[f\left( x \right) = \left\{ {\begin{array}{rc}{7 - 4x}&{x < 1}\\{{x^2} + 2}&{x \ge 1}\end{array}} \right.\]

Evaluate the following limits, if they exist.

  1. \(\mathop {\lim }\limits_{x \to \, - 6} f\left( x \right)\)
  2. \(\mathop {\lim }\limits_{x \to 1} f\left( x \right)\)
Hint : Recall that when looking at overall limits (as opposed to one-sided limits) we need to make sure that the value of the function must be approaching the same value from both sides. In other words, the two one sided limits must both exist and be equal.

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a \(\mathop {\lim }\limits_{x \to \, - 6} f\left( x \right)\) Show Solution

For this part we know that \( - 6 < 1\) and so there will be values of x on both sides of -6 in the range \(x < 1\) and so we can assume that, in the limit, we will have \(x < 1\). This will allow us to use the piece of the function in that range and then just use standard limit techniques to compute the limit.

\[\mathop {\lim }\limits_{x \to \, - 6} f\left( x \right) = \mathop {\lim }\limits_{x \to \, - 6} \left( {7 - 4x} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{31}}\]

b \(\mathop {\lim }\limits_{x \to 1} f\left( x \right)\) Show Solution

This part is going to be different from the previous part. We are looking at the limit at \(x = 1\) and that is the “cut–off” point in the piecewise functions. Recall from the discussion in the section, that this means that we are going to have to look at the two one sided limits.

\[\mathop {\lim }\limits_{x \to {1^{\, - }}} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^{\, - }}} \left( {7 - 4x} \right) = \underline 3 \hspace{0.25in}\hspace{0.25in}{\mbox{because }}x \to {1^ - }{\mbox{ implies that }}x < 1\] \[\mathop {\lim }\limits_{x \to {1^{\, + }}} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^{\, + }}} \left( {{x^2} + 2} \right) = \underline 3 \hspace{0.25in}\hspace{0.25in}{\mbox{because }}x \to {1^ + }{\mbox{ implies that }}x > 1\]

So, in this case, we can see that,

\[\mathop {\lim }\limits_{x \to {1^{\, - }}} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^{\, + }}} f\left( x \right) = 3\]

and so we know that the overall limit must exist and,

\[\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \require{bbox} \bbox[2pt,border:1px solid black]{3}\]