Respuesta :

Explanation

So we must find the end behaviour of the following function when x tends to infinite:

[tex]f(x)=3^{-x}+2[/tex]

This basically means that we must find the following limit:

[tex]\lim_{x\to\infty}f(x)=\lim_{x\to\infty}(3^{-x}+2)[/tex]

Before continuing it would be a good idea to remember the following property of powers:

[tex]a^{-n}=\frac{1}{a^n}[/tex]

So if we apply this property to f(x) we get:

[tex]f(x)=\frac{1}{3^x}+2[/tex]

So the limit that we have to calculate is:

[tex]\lim_{x\to\infty}f(x)=\lim_{x\to\infty}(\frac{1}{3^x}+2)[/tex]

When x tends to infinite the exponential term 3ˣ also tends to infinite. This means that when x tends to infinite the term 1/3ˣ tends to 1 divided by infinite. A constant divided by an expression tending to infinite always tends to 0 since you are dividing a fixed number by another that keeps increasing. Then the limit above is:

[tex]\begin{gathered} \lim_{x\to\infty}(\frac{1}{3^x}+2)=(0+2)=2 \\ \lim_{x\to\infty}f(x)=2 \end{gathered}[/tex]Answer

Then the answer is that f(x) tends to 2 when x tends to infinite.

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