Question

1. evaluate the following limits:

a) $lim_{x
ightarrowinfty}\frac{x^{7}+8x + 5}{x^{8}+9x + 10}$
b) $lim_{x
ightarrowinfty}\frac{sin(8x)}{8x}$

Explanation:

Step1: Divide numerator and denominator by highest - power of x

For $\lim_{x
ightarrow\infty}\frac{x^{7}+8x + 5}{x^{8}+9x + 10}$, divide both numerator and denominator by $x^{8}$:
$\lim_{x
ightarrow\infty}\frac{\frac{x^{7}}{x^{8}}+\frac{8x}{x^{8}}+\frac{5}{x^{8}}}{\frac{x^{8}}{x^{8}}+\frac{9x}{x^{8}}+\frac{10}{x^{8}}}=\lim_{x
ightarrow\infty}\frac{\frac{1}{x}+\frac{8}{x^{7}}+\frac{5}{x^{8}}}{1+\frac{9}{x^{7}}+\frac{10}{x^{8}}}$

Step2: Evaluate the limit

As $x
ightarrow\infty$, $\frac{1}{x}
ightarrow0$, $\frac{8}{x^{7}}
ightarrow0$, $\frac{5}{x^{8}}
ightarrow0$, $\frac{9}{x^{7}}
ightarrow0$ and $\frac{10}{x^{8}}
ightarrow0$.
So, $\lim_{x
ightarrow\infty}\frac{\frac{1}{x}+\frac{8}{x^{7}}+\frac{5}{x^{8}}}{1+\frac{9}{x^{7}}+\frac{10}{x^{8}}}=0$

Step3: Analyze the second limit

For $\lim_{x
ightarrow\infty}\frac{\sin(8x)}{8x}$, we know that $- 1\leqslant\sin(8x)\leqslant1$.
So, $-\frac{1}{8x}\leqslant\frac{\sin(8x)}{8x}\leqslant\frac{1}{8x}$

Step4: Evaluate the limit using Squeeze - Theorem

As $x
ightarrow\infty$, $\lim_{x
ightarrow\infty}-\frac{1}{8x}=0$ and $\lim_{x
ightarrow\infty}\frac{1}{8x}=0$
By the Squeeze - Theorem, $\lim_{x
ightarrow\infty}\frac{\sin(8x)}{8x}=0$

Answer:

a) $0$
b) $0$