Answer:
A
Step-by-step explanation:
We are given the function:
[tex]\displaystyle f(x) = \left\{ \begin{array}{ll} 2\cos(\pi x) \text{ for } x \leq -1 \\ \\ \displaystyle \frac{2}{\cos(\pi x)}\text{ for } x > -1 \end{array} \right.[/tex]
And we want to find:
[tex]\displaystyle \lim_{x\to -1}f(x)[/tex]
So, we need to determine whether or not the limit exists. In other words, we will find the two one-sided limits.
Left-Hand Limit:
[tex]\displaystyle \lim_{x\to-1^-}f(x)[/tex]
Since we are approaching from the left, we will use the first equation:
[tex]\displaystyle =\lim_{x\to -1^-}2\cos(\pi x)[/tex]
By direct substitution:
[tex]=2\cos(\pi (-1))=2\cos(-\pi)=2(-1)=-2[/tex]
Right-Hand Limit:
[tex]\displaystyle \lim_{x\to -1^+}f(x)[/tex]
Since we are approaching from the right, we will use the second equation:
[tex]=\displaystyle \lim_{x\to -1^+}\frac{2}{\cos(\pi x)}[/tex]
Direct substitution:
[tex]\displaystyle =\frac{2}{\cos(\pi (-1))}=\frac{2}{\cos(-\pi)}=\frac{2}{(-1)}=-2[/tex]
So, we can see that:
[tex]\displaystyle \displaystyle \lim_{x\to-1^-}f(x)=\displaystyle \lim_{x\to -1^+}f(x) =-2[/tex]
Since both the left- and right-hand limits exist and equal the same thing, we can conclude that:
[tex]\displaystyle \lim_{x \to -1}f(x)=-2[/tex]
Our answer is A.
