Answer:
1
Step-by-step explanation:
To solve the given limit using the Squeeze Theorem (or Pinch Theorem), we need to find two functions that "squeeze" our given function. In this case, we can exploit the fact that the sine function oscillates between -1 and 1, regardless of the value of 'x'.
We are given:
[tex]\lim_{x \to \infty} \left(1 + e^{-3x}\sin(x)\right)[/tex]
Using the Squeeze Theorem below:
[tex]\boxed{\begin{minipage}{8 cm}\underline{Squeeze Theorem:}\\\\If $g(x) \leq f(x) \leq h(x)$ near $a$, and $\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L,$ then $\lim_{x \to a} f(x) = L.$\\\\where:\\\phantom{ww}$\bullet$ $f(x)$ is squeezed by $g(x)$ and $h(x)$.\\ \phantom{ww}$\bullet$ $L$ is the common limit of $g(x)$ and $h(x)$ as $x$ approaches $a$.\end{minipage}}[/tex]
And knowing that -1 ≤ sin(x) ≤ 1, we can say:
[tex]\lim_{x \to \infty} \left(1 + e^{-3x}(-1)\right) \leq \lim_{x \to \infty} \left(1 + e^{-3x}\sin(x)\right) \leq \lim_{x \to \infty} \left(1 + e^{-3x}(1)\right)[/tex]
[tex]\Longrightarrow \lim_{x \to \infty} \left(1 - e^{-3x}\right) \leq \lim_{x \to \infty} \left(1 + e^{-3x}\sin(x)\right) \leq \lim_{x \to \infty} \left(1 + e^{-3x}\right)[/tex]
Now, we examine the limits of the bounding functions as 'x' approaches infinity:
[tex]\Longrightarrow \lim_{x \to \infty} \left(1 - e^{-\infty}\right) \leq \lim_{x \to \infty} \left(1 + e^{-3x}\sin(x)\right) \leq \lim_{x \to \infty} \left(1 + e^{-\infty}\right)[/tex]
[tex]\Longrightarrow 1 \leq \lim_{x \to \infty} \left(1 + e^{-3x}\sin(x)\right) \leq 1[/tex]
Since the limits of the bounding functions approach 1, the given limit must also approach 1.
