Question

suppose we wish to evaluate $lim_{x
ightarrow-infty}15e^{9x}cos(7x)+7$. since we know that $- 1leqcos(\theta)leq1$ for $\thetainmathbb{r}$, we must have $-1leqcos(7x)leq1$ for $x
eq0$. from this, we must have $-15e^{9x}leq15e^{9x}cos(7x)leq15e^{9x}$ for $x
eq0$. $-15e^{9x}+7leq15e^{9x}cos(7x)+7leq15e^{9x}+7$ for $x
eq0$. evaluate the limits. $lim_{x
ightarrow-infty}-15e^{9x}+7=-infty$ $lim_{x
ightarrow-infty}15e^{9x}+7=-infty$

Explanation:

Step1: Recall exponential - limit property

The limit of $e^{ax}$ as $x\to-\infty$ for $a > 0$ is 0. Here $a = 9>0$.

Step2: Evaluate left - hand side limit

We want to find $\lim_{x\to-\infty}(-15e^{9x}+7)$. Using the limit property $\lim_{x\to-\infty}e^{9x}=0$, we have $\lim_{x\to-\infty}(-15e^{9x}+7)=-15\times0 + 7=7$.

Step3: Evaluate right - hand side limit

We want to find $\lim_{x\to-\infty}(15e^{9x}+7)$. Using the limit property $\lim_{x\to-\infty}e^{9x}=0$, we have $\lim_{x\to-\infty}(15e^{9x}+7)=15\times0 + 7=7$.

Step4: Apply Squeeze Theorem

Since $-15e^{9x}+7\leq15e^{9x}\cos(7x)+7\leq15e^{9x}+7$ for $x
eq0$ and $\lim_{x\to-\infty}(-15e^{9x}+7)=\lim_{x\to-\infty}(15e^{9x}+7) = 7$, by the Squeeze Theorem, $\lim_{x\to-\infty}(15e^{9x}\cos(7x)+7)=7$.

Answer:

$7$