Question

find the limit.
lim_{x
ightarrow-infty}\frac{cos 8x}{7x}
lim_{x
ightarrow-infty}\frac{cos 8x}{7x}=square \text{ (simplify your answer.)}

Explanation:

Step1: Recall cosine - function range

The range of $\cos(8x)$ is $[- 1,1]$, i.e., $-1\leqslant\cos(8x)\leqslant1$.

Step2: Analyze the limit

We have $\lim_{x
ightarrow-\infty}\frac{\cos(8x)}{7x}$. Divide the inequality of $\cos(8x)$ by $7x$ (since $x
ightarrow-\infty$, $7x<0$). So $\frac{1}{7x}\leqslant\frac{\cos(8x)}{7x}\leqslant\frac{-1}{7x}$.

Step3: Find the limits of the bounds

$\lim_{x
ightarrow-\infty}\frac{1}{7x}=0$ and $\lim_{x
ightarrow-\infty}\frac{-1}{7x}=0$.

Step4: Apply the Squeeze - Theorem

By the Squeeze - Theorem, if $f(x)\leqslant g(x)\leqslant h(x)$ for all $x$ in some open interval containing $a$ (in this case as $x
ightarrow-\infty$) and $\lim_{x
ightarrow-\infty}f(x)=\lim_{x
ightarrow-\infty}h(x) = L$, then $\lim_{x
ightarrow-\infty}g(x)=L$. Here, $f(x)=\frac{1}{7x}$, $g(x)=\frac{\cos(8x)}{7x}$, $h(x)=\frac{-1}{7x}$ and $L = 0$.

Answer:

$0$