Question

1. (7 points) find the limit. for (b)-(d), fully justify your answers.

a. (1 point) given that $lim_{x
ightarrow1}f(x)=2$, $lim_{x
ightarrow1}g(x)=3$, and $lim_{x
ightarrow1}h(x)=5$, what is $lim_{x
ightarrow1}\frac{f(x)}{g(x)h(x)}=$
b. (2 points) $lim_{x
ightarrowinfty}\frac{5x^{2}+e^{-x}}{7x^{2}-3}=$
c. (2 points) $lim_{x
ightarrowinfty}\frac{2x^{3}+e^{x}}{5x^{3}-4}=$
d. (2 points) $lim_{x
ightarrowinfty}\frac{x^{2}-9}{x^{2}-5x + 6}=$

Explanation:

Step1: Use limit - quotient and product rules for part a

By the quotient rule $\lim_{x
ightarrow a}\frac{u(x)}{v(x)}=\frac{\lim_{x
ightarrow a}u(x)}{\lim_{x
ightarrow a}v(x)}$ and product rule $\lim_{x
ightarrow a}(u(x)v(x))=\lim_{x
ightarrow a}u(x)\cdot\lim_{x
ightarrow a}v(x)$ where $\lim_{x
ightarrow a}v(x)
eq0$.
So, $\lim_{x
ightarrow1}\frac{f(x)}{g(x)h(x)}=\frac{\lim_{x
ightarrow1}f(x)}{\lim_{x
ightarrow1}g(x)\cdot\lim_{x
ightarrow1}h(x)}$.
Substitute $\lim_{x
ightarrow1}f(x) = 2$, $\lim_{x
ightarrow1}g(x)=3$ and $\lim_{x
ightarrow1}h(x)=5$:
$\frac{2}{3\times5}=\frac{2}{15}$.

Step2: For part b, divide numerator and denominator by $x^{2}$

$\lim_{x
ightarrow\infty}\frac{5x^{2}+e^{-x}}{7x^{2}-3}=\lim_{x
ightarrow\infty}\frac{5 + \frac{e^{-x}}{x^{2}}}{7-\frac{3}{x^{2}}}$.
Since $\lim_{x
ightarrow\infty}\frac{e^{-x}}{x^{2}} = 0$ (using L - H rule or the fact that exponential decay $e^{-x}$ is much faster than polynomial growth $x^{2}$ as $x
ightarrow\infty$) and $\lim_{x
ightarrow\infty}\frac{3}{x^{2}}=0$.
The limit is $\frac{5 + 0}{7-0}=\frac{5}{7}$.

Step3: For part c, divide numerator and denominator by $x^{3}$

$\lim_{x
ightarrow\infty}\frac{2x^{3}+e^{x}}{5x^{3}-4}=\lim_{x
ightarrow\infty}\frac{2+\frac{e^{x}}{x^{3}}}{5 - \frac{4}{x^{3}}}$.
As $x
ightarrow\infty$, $\lim_{x
ightarrow\infty}\frac{e^{x}}{x^{3}}=\infty$ (using L - H rule multiple times: $\lim_{x
ightarrow\infty}\frac{e^{x}}{x^{3}}=\lim_{x
ightarrow\infty}\frac{e^{x}}{3x^{2}}=\lim_{x
ightarrow\infty}\frac{e^{x}}{6x}=\lim_{x
ightarrow\infty}\frac{e^{x}}{6}=\infty$). So, $\lim_{x
ightarrow\infty}\frac{2x^{3}+e^{x}}{5x^{3}-4}=\infty$.

Step4: For part d, divide numerator and denominator by $x^{2}$

$\lim_{x
ightarrow\infty}\frac{x^{2}-9}{x^{2}-5x + 6}=\lim_{x
ightarrow\infty}\frac{1-\frac{9}{x^{2}}}{1-\frac{5}{x}+\frac{6}{x^{2}}}$.
Since $\lim_{x
ightarrow\infty}\frac{9}{x^{2}} = 0$, $\lim_{x
ightarrow\infty}\frac{5}{x}=0$ and $\lim_{x
ightarrow\infty}\frac{6}{x^{2}}=0$.
The limit is $\frac{1-0}{1 - 0+0}=1$.

Answer:

a. $\frac{2}{15}$
b. $\frac{5}{7}$
c. $\infty$
d. $1$