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anastassius [24]
3 years ago
6

Given f(x) =

Mathematics
1 answer:
sergejj [24]3 years ago
8 0

Answer:

A

Step-by-step explanation:

We are given the function:

\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.

And we want to find:

\displaystyle \lim_{x\to -1}f(x)

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:

\displaystyle \lim_{x\to-1^-}f(x)

Since we are approaching from the left, we will use the first equation:

\displaystyle =\lim_{x\to -1^-}2\cos(\pi x)

By direct substitution:

=2\cos(\pi (-1))=2\cos(-\pi)=2(-1)=-2

Right-Hand Limit:

\displaystyle \lim_{x\to -1^+}f(x)

Since we are approaching from the right, we will use the second equation:

=\displaystyle \lim_{x\to -1^+}\frac{2}{\cos(\pi x)}

Direct substitution:

\displaystyle =\frac{2}{\cos(\pi (-1))}=\frac{2}{\cos(-\pi)}=\frac{2}{(-1)}=-2

So, we can see that:

\displaystyle \displaystyle \lim_{x\to-1^-}f(x)=\displaystyle \lim_{x\to -1^+}f(x) =-2

Since both the left- and right-hand limits exist and equal the same thing, we can conclude that:

\displaystyle \lim_{x \to -1}f(x)=-2

Our answer is A.

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