Q. 1 - Calculus

Question

Determine whether each of the statements that follow is true or false. If a statement is true, explain why. If a statement is false, provide a counterexample.

(a) True or False: If $f\left(x\right)\to 0^{+},$ then $\frac{1}{f\left(x\right)}\to \infty .$

(b) True or False: If $f\left(x\right)\to {\infty }^{+},$ then $\frac{1}{f\left(x\right)}\to 0^{+}.$

(c) True or False: If a limit initially has an indeterminate form, then it can never be solved. 

(d) True or False: A limit “does not exist” if there is no real number that it approaches.

(e) True or False: As limit forms, ${\infty }^2\to \infty .$

(f) True or False: As limit forms, $2^{\infty }\to \infty .$

(g) True or False: As limit forms, $\infty -\infty \to 0$

(h) True or False: The limit of a function $f$ as $x\to c$ is always equal value $f\left(c\right),$ provided that $f\left(c\right)$exists.

Step-by-Step Solution

Answer

All the given statements are true.

Part (a) Step 1. Explanation

It is known that,

If $\underset{x\to c}{\lim }\frac{g\left(x\right)}{f\left(x\right)}$ is of the form $\frac{1}{0^{+}}$ then $\underset{x\to c}{\lim }\frac{g\left(x\right)}{f\left(x\right)}=\infty$.

Now for this case assume $f\left(x\right)\to 0^{+} \& g\left(x\right)\to 1.$

Hence, the given statement is true.

Part (b) Step 1. Explanation

It is known that,

If $\underset{x\to c}{\lim }\frac{g\left(x\right)}{f\left(x\right)}$ is of the form $\frac{1}{\infty } then \underset{x\to c}{\lim }\frac{g\left(x\right)}{f\left(x\right)}=0.$

Now for this case assume $f\left(x\right)\to {\infty }^{+} \& g\left(x\right)\to 0^{+}.$

Hence, the given statement is true.

Part (c) Step 1. Explanation

Given statement is false. Counter example for the statement is given below.

$$\begin{gathered} \underset{x\to 2}{\lim }\frac{x^2-4}{x-2} \\ \underset{x\to 2}{\lim }\frac{x^2-4}{x-2}=\underset{x\to 2}{\lim }\frac{(x-2)(x+2)}{x-2} \\ =\underset{x\to 2}{\lim }(x+2) \\ =4 \end{gathered}$$

Part (d) Step 1. Explanation

Given statement is false. Counter example for the statement is given below.

$\underset{x\to \infty }{\lim }\left(\ln \left(x\right)\right)=\infty$

Part (e) Step 1. Explanation

Since $\infty$ is value which is not known therefore any power of $\infty$i.e. not known number.

Part(f) Step 1. Explanation

Since $\infty$is value which is not known therefore $2\times 2\times 2\times 2\times .......{\times }_{\infty times}2$ will also lead to $\infty$ i.e. an unknown number.

Part (g) Step 1. Explanation

It is already known  that $\infty -\infty$ is an indeterminate form and thus its value cannot be equal to zero.

Part (h) Step 1. Explanation

Given statement is false. Counter example  for the statement is given below.

$$\begin{gathered} f\left(x\right)=\left\{\begin{array}{l}10:x=2 \\ x+1:0<x<2 \\ 2x-1:2<x<5\\ \end{array}\right. \end{gathered}$$

Left hand limit $=$right hand limit $=4$

$f\left(2\right)=10$