First Principles AI

Question 1

Determine the left hand limit and the right hand limit and then find limit of the following functions when $x \to c$ .

(i)

$f(x) = 2x^2 + x - 5$ , $c = 1$

(ii)

$f(x) = \dfrac{x^2 - 9}{x - 3}$ , $c = -3$

(iii)

$f(x) = |x - 5|$ , $c = 5$

Solution Theory

Question 2

Discuss the continuity of $f(x)$ at $x = c$

(i)

$f(x) = \begin{cases} 2x + 5 & \text{if } x \le 2 \\ 4x + 1 & \text{if } x > 2 \end{cases}$ , $c = 2$

(ii)

$f(x) = \begin{cases} 3x - 1 & \text{if } x < 1 \\ 4 & \text{if } x = 1 \\ 2x & \text{if } x > 1 \end{cases}$ , $c = 1$

Solution Theory

Question 3

If $f(x) = \begin{cases} 3x & \text{if } x \le -2 \\ x^2 - 1 & \text{if } -2 < x < 2 \\ 3 & \text{if } x \ge 2 \end{cases}$ . Discuss continuity at $x = 2$ and $x = -2$ .

Solution Theory

Question 4

If $f(x) = \begin{cases} x + 2 & x \le -1 \\ c + 2 & x > -1 \end{cases}$ , find “ $c$ ” so that $\lim_{x \to -1} f(x)$ exists.

Solution Theory

Question 5

Find the values of $m$ and $n$ , so that given function $f$ is continuous at $x = 3$

(i)

$f(x) = \begin{cases} mx & \text{if } x < 3 \\ n & \text{if } x = 3 \\ -2x + 9 & \text{if } x > 3 \end{cases}$

(ii)

$f(x) = \begin{cases} mx & \text{if } x < 3 \\ x^2 & \text{if } x \ge 3 \end{cases}$

Solution Theory

Question 6

$f(x) = \begin{cases} \dfrac{\sqrt{2x + 5} - \sqrt{x + 7}}{x - 2} & , x \ne 2 \\ k & , x = 2 \end{cases}$ . Find value of $k$ so that $f$ is continuous at $x = 2$ .

Solution Theory

Question 7

Given the function $f(x) = \begin{cases} 2x + 3, & x \le 1 \\ -x + 4, & x > 1 \end{cases}$ . Discuss the limit and continuity at $x = 1$ .

Solution Theory

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