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

$\begin{aligned} \therefore \quad \lim _{x \rightarrow+\infty} \frac{5 x^2+7}{3 x^2-x} & =\lim _{x \rightarrow+\infty} \frac{5 x^2}{3 x^2} \\ & =\lim _{x \rightarrow+\infty} \frac{5 }{3}=5 / 3 \end{aligned}$