Problem

Sketch the graph of the function and use it to de…

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Problem 11 Medium Difficulty

Sketch the graph of the function and use it to determine the values of $ a $ for which $ \displaystyle \lim_{x\to a}f(x) $ exists.
$$ f(x) = \left\{
\begin{array}{ll}
1 + x & \mbox {if $ x < -1 $}\\
x^2 & \mbox{if $ -1 \le x < 1$}\\
2 - x & \mbox{if $ x \ge 1 $}
\end{array} \right.$$

Answer

From the graph of
\[.f(x)=\left\{\begin{array}{ll}1+x & \text { if } x<-1 \\x^{2} & \text { if }-1 \leq x<1 \\2-x & \text { if } x \geq 1\end{array}\right.\]
we see that $\lim _{x \rightarrow a} f(x)$ exists for all $a$ except $a=-1 .$ Notice that the
right and left limits are different at $a=-1$

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Video Transcript

defined in three different parts. Uh function is defined to be one plus X. When x is less than negative one, that's the red portion of the graph that you see. The function f of X equals X squared. When uh negative one is less than or equal to ex uh ex in turn is strictly less than one. That's the blue portion of the graph that you see. And last but not least when x is greater than or equal to one. Uh F of X is defined to be two minus X. We want to find all values a along the X axis for which the limit exists. And this graph is uh going to greatly help us. You can see um that this uh function is basically connected. There's no breaks anywhere in the function except right here. When X is negative one, we have a break into function when X is positive one. Uh You can see the blue portion of the function uh meets nicely uh the green portion of the function. And so the limit as X approaches one from the negative side, the blue side is going to be the same limit as X approaches one from the uh positive side. The green side. But over here, it when uh X is negative one, or specifically when a is equal to negative one. Uh This function uh does not have a limit in other words. Uh The limit of f of X as X approaches negative one does not exist. Um The limits exist for dysfunction for any other value, any other real number value along the X axis, it does not exist. The limit of dysfunction does not exist when X approaches the value of negative one because as X approaches negative one from the negative side to function uh approaches zero. So the limit of F of X as X approaches negative one from the negative side is zero, but the limit of F of X pay attention to the blue portion of the curve. Now the limit of F of X as X approaches negative one from the positive side, uh F of X is approaching the value of one. And so the limit of F of X as X approaches negative one does not exist because the limit of F of X as X approaches negative one from the negative side does not equal the limit of F. Of X as X approaches negative one from the positive side. And you can clearly see that because the red portion of the graph is getting close to zero, think of the Y axis as to function values, whereas the blue portion of the graph is getting close to the value of one

Leon D.

Temple University

Related Courses

Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 2

Limits and Derivatives

Section 2

The Limit of a Function

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