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Math 181 Miniproject 3: Texting Lesson.md
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My lesson Topic
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<div><img class="left"/><div class="alert gray">

Spongebob! Have you started the review for the first calculus exam yet? I need help with Problem Type #0: Determining Limits, Continuity, and Differentiability Using Only a Graph. The exam is tomorrow! Help! -Patrick 

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Given $f(x):$

![](https://i.imgur.com/fLFENEf.png)

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a) $\lim_{h \to 0}f(x)$

b) $\lim_{h \to -2}f(x)$

c) Identify all values such that $\lim_{h \to a}f(x)$ is undefined.

d) Identify all values such that $f(x)$ is undefined at $x=a$.

e) Identify all values of $a$ such that $f(x)$ is discontinuous at $x=a$.

f) Identify all values of $a$ such that $f(x)$ is not differentiable at $x=a$.

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In lecture the professor commented that some conditions or statements about a function are "stronger" than others. What does that mean in the context of this problem and how can I use that information in relation to determining the limit, continuity, or differentiability of a function? 

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Hi Patrick! 

It is important to be able to determine limits, continuity, and differentiability of a function graphically because you may not always be given the function directly. It is also important to understand what it means for a limit to exist, for a function to be continuous, and for a function to be differentiable and these concepts can all be easily illustrated by using a graph as a model. Before I answer your question, we let's talk about how to solve some of the problem types first.

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<div><div class="alert blue">

 To do parts a) and b): we are being asked to find the limit of $f(x)$ as our input values approach $x=-2$ and $x=0$. To do this, we determine the output $(y)$ values that $f(x)$ seems to be getting closer to as our input $(x)$ values get closer to -2 and 0, from both sides of the function when not specified otherwise. In order for the limit to exist, the same output value must be approached by the function from both sides. Therefore:

a) $\lim_{h \to 0}f(x)$=5

b) $\lim_{h \to -2}f(x)$= does not exist

The limit as $f(x)$ approaches $x=-2$ does not exist since both sides of the function approach different output values (+3 from the right and +1 from the left). Remember that the function does not necessarily have to exist at the x value in question in order to have a limit; it just needs to "approach" the same output value from either side. This information can also help us figure out part c); that is, the limit will not exist if $f(x)$ approaches differing output values from either side of the function. Therefore:
The limit will not exist as $f(x)$ approaches $x=-2$ and $x=+3$. 


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Remember for the exam, in cases of the function having a vertical asymptote, the limit will not exist as the function approaches this x value since both sides will be approaching different output values. 
This wasn't shown in this review but is a good thing to be aware of. 

![](https://i.imgur.com/9LrIKlV.jpg)


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For part d): the function appears to be defined at every point shown on the graph, even though there are "open circles" at certain points. To determine this, ask yourself if there is an existing output $(y)$ value for the input $(x)$ value in question. This relates to part e) in that if a function is not defined at any specific point, it will not be continuous there. For a function to be continuous: the limit at the input value of interest must exist and it must be equal to the function's output value at that point. 
Therefore: 
The function is not continuous at $x=-2$ and $x=+3$ because the limit does not exist at either of these two points. 

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Is that what the professor is referring to when they say some conditions are "stronger" than others? 

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Yes, in order for the function to be continuous at a point, the "weaker" statement (that the limit existed at this point) had to be true. This will also help us in part f). By this, I mean that in order for a function to be differentiable, the two preceding statements must be true: the limit must exist and the function has to be continuous at the point of interest in order for it to be differentiable there. Being differentiable is the strongest statement; if the function is differentiable at a point then that means the limit exists and it is continuous at that point as well. 

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So does this mean that as long as the limit exists and the function is continuous, the function should also be differentiable?

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Be careful because there are some instances where the limit exists and the function is continuous at the point of interest but the function is still not differentiable at this point. For example, in this problem the function cannot be differentiated at $x=0$ because there is a sharp change in slope at this point and therefore the slope at this point is undefined. There are other common instances where this is possible so you should watch out for those on the exam tomorrow! Examples of these include any graphs that have rapid changes in slope (like a sharp corner) or any point where the slope becomes undefined (like a vertical line). I will send pictures of examples. 

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![](https://i.imgur.com/ZNRsWGz.jpg)


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![](https://i.imgur.com/8oec7rC.jpg)


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Thank you for all of your help Spongebob! 

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You're welcome Patrick but in the future, try not to wait until the last minute to study calculus or ask for help! This doesn't have to be you:
![](https://i.imgur.com/kl47dne.jpg)


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