Calculus Solutions Ex#1.3

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Q.09
Solution:
$$\lim_{x \rightarrow+\infty}\left(1+2 x-3 x^5\right)=\lim_{x \rightarrow+\infty}\left(-3 x^5\right)=-\infty$$
This is because the behavior of a polynomial matches the end behavior of its highest degree term.
Q.10
Solution:
$$\lim _{x \rightarrow+\infty}\left(2 x^3-100 x+5\right)=\lim_{x \rightarrow+\infty} 2 x^3=+\infty.$$ $ \quad \left(2 x^3\right.$ is the highest degree term).
Q.11
Solution:
$$\quad \lim_{x \rightarrow+\infty} \sqrt{x}=\sqrt{\lim_{x \rightarrow+\infty} x}=+\infty$$.
Q.12
Solution:
$$\lim _{x \rightarrow-\infty} \sqrt{5-x}=\sqrt{\lim _{x \rightarrow-\infty} (5-x)}=\sqrt{5+\infty}=+\infty$$
Q.13
$$\lim _{x \rightarrow+\infty} \frac{3 x+1}{2 x-5}$$
Solution:
$3x$ is the highest degree term in the numerater and $2x$ is the highest degree term in denominator $$\therefore \lim _{x \rightarrow+\infty} \frac{3 x+1}{2 x-5}=\lim _{x \rightarrow+\infty} \frac{3 x}{2 x}=\lim _{x \rightarrow+\infty} 3 / 2=3/2$$.
Q.14
$$\lim _{x \rightarrow+\infty} \frac{5 x^2-4 x}{2 x^2+3}$$
Solution:
$$\lim _{x \rightarrow+\infty} \frac{5 x^2-4 x}{2 x^2+3}$$ $5 x^2$ is the highest degree term in the numerator and $2 x^2$ is the highest degree term in the denominator$$ \therefore \quad \lim _{x \rightarrow+\infty} \frac{5 x^2-4 x}{2 x^2+3}=\lim _{x \rightarrow+\infty} \frac{5 x^2}{2 x^2}=\lim _{x \rightarrow+\infty} 5 / 2=5 / 2$$
Q.15
$$\lim _{y \rightarrow-\infty} \frac{3}{y+4}$$
Solution:
$$\lim _{y \rightarrow-\infty} \frac{3}{y+4}=\lim _{y \rightarrow-\infty} \frac{3}{y}=0$$
because as $y \rightarrow-\infty, \quad  \frac{1}{y} \rightarrow 0$
Q.16
$$\lim _{x \rightarrow+\infty} \frac{1}{x-12}$$
Solution:
$$\begin{aligned} & \lim _{x \rightarrow+\infty} \frac{1}{x-12}=\lim _{x \rightarrow+\infty} \frac{1}{x}=0 \\ & \therefore \text { as } x \rightarrow+\infty, \frac{1}{x} \rightarrow 0 . \end{aligned}$$
Q.17
Solution:
$$\lim _{x \rightarrow-\infty} \frac{x-2}{x^2+2 x+1}$$
$x$ is the highest degree term in the numerator and $x^2$ is the highest degree term in the denominator. $$\begin{aligned} & \therefore \lim _{x \rightarrow-2} \frac{x-2}{x^2+2 x+1}=\lim _{x \rightarrow-\infty} \frac{x}{x^2} \\ &=\lim _{x \rightarrow-\infty} \frac{1}{x}=0 \\ & \therefore \quad as, x \rightarrow-\infty, \quad \frac{1}{x} \rightarrow 0 \end{aligned}$$
Q.18
Solution:
$$\lim _{x \rightarrow+\infty} \frac{5 x^2+7}{3 x^2-x}$$ $5 x^2$ and $3 x^2$ are the highest degree terms in the numerator and denominator respectively $$\begin{aligned} \therefore \quad \lim _{x \rightarrow+\infty} \frac{5 x^2+7}{3 x^2-x} & =\lim _{x \rightarrow+\infty} \frac{5 x^2}{3 x^2} \\ & =\lim _{x \rightarrow+\infty} \frac{5 }{3}=5 / 3 \end{aligned}$$
Q.19
$$\lim _{x \rightarrow+\infty} \frac{7-6 x^5}{x+3}$$
Solution:
$$\lim _{x \rightarrow+\infty} \frac{7-6 x^5}{x+3}=\lim _{x \rightarrow+\infty} \frac{-6 x^5}{x}$$
Because $-6 x^5$ and $x$ are the highest-degree terms in the numerator and denominator, respectively. Therefore, $$\lim _{x \rightarrow+\infty} \frac{-6 x^5}{x}=\lim _{x \rightarrow+\infty}-6 x^4=-\infty$$