Chapter 11.1, Problem 77E

(a)

To determine

To graph: The functionand explain whether the limits can be reached or not.

Answer to Problem 77E

Here, $\underset{x\to 1}{\lim }2x^2+x-4$ and $\underset{x\to 3}{\lim }\frac{x-3}{x^2-9}$ can be reached.

Explanation of Solution

Given: $\underset{x\to 1}{\lim }2x^2+x-4$ and $\underset{x\to 3}{\lim }\frac{x-3}{x^2-9}$

Calculation:

Here, graph the functions from exercise 4 and 5. The limit from exercise 4 can be reached, since the function is just a polynomial function. The limit from exercise 5 cannot be reached, since the function causes a division by 0 when $x=3$ . However, from looking at both graphs in the standard viewing window, it seems as though both limits can be reached, since the graph look continuous around where the limits are evaluated(the function in exercise 5 is undefined only $x$ is equal to exactly 3, and the graph looks continuous and smooth otherwise).

Graph of $\underset{x\to 1}{\lim }2x^2+x-4$ from exercise 4:

  

Graph of $\underset{x\to 3}{\lim }\frac{x-3}{x^2-9}$

from exercise 5:

Conclusion:

Hence, $\underset{x\to 1}{\lim }2x^2+x-4$ and $\underset{x\to 3}{\lim }\frac{x-3}{x^2-9}$ can be reached.

(b)

To determine

To graph: The functionand explain whether the limits can be reached or not.

Answer to Problem 77E

Here, $\underset{x\to 1}{\lim }2x^2+x-4$ is reachable and $\underset{x\to 3}{\lim }\frac{x-3}{x^2-9}$ is unreachable.

Explanation of Solution

Calculation:

Here, graph both functions again the decimal setting and zooming in more closely on the values to which the limits are being taken. Now, see that while the limit from exercise 4 is reachable and the function is continuous around this limit, the one from exercise 5 is unreachable, and the function has a gap at that value.

  

Graph of $\underset{x\to 1}{\lim }2x^2+x-4$ from exercise 4:

  

Graph of $\underset{x\to 3}{\lim }\frac{x-3}{x^2-9}$ from exercise 5:

  

Conclusion:

Hence, $\underset{x\to 1}{\lim }2x^2+x-4$ is reachable and $\underset{x\to 3}{\lim }\frac{x-3}{x^2-9}$ is unreachable.