Question

1. (6 marks) evaluate the following limits:

(a) $lim_{x \to 3} (2x + 5)$
(b) $lim_{x \to -1} (x^2 + 4x + 3)$
(c) $lim_{x \to 2} \frac{x^2 - 4}{x - 2}$

1. (6 marks) consider the function

$f(x)=\

$$\begin{cases}x^2, & x \ eq 1 \\\\3, & x = 1\\end{cases}$$

$
error! filename not specified.
(a) find $lim_{x \to 1} f(x)$.
(b) find $f(1)$.
(c) is the function continuous at $x = 1$? justify your answer.

Explanation:

Problem 9

(a)

Step1: Substitute $x=3$

$\lim_{x \to 3} (2x+5) = 2(3) + 5$

Step2: Calculate the result

$2(3)+5 = 6+5=11$

(b)

Step1: Substitute $x=-1$

$\lim_{x \to -1} (x^2+4x+3) = (-1)^2 + 4(-1) + 3$

Step2: Calculate the result

$1 - 4 + 3 = 0$

(c)

Step1: Factor numerator

$x^2-4=(x-2)(x+2)$, so $\lim_{x \to 2} \frac{(x-2)(x+2)}{x-2}$

Step2: Cancel common factor

$\lim_{x \to 2} (x+2)$ (since $x
eq 2$)

Step3: Substitute $x=2$

$2+2=4$

(a)

Step1: Use $x^2$ for $x

eq 1$
$\lim_{x \to 1} f(x) = \lim_{x \to 1} x^2$

Step2: Substitute $x=1$

$\lim_{x \to 1} x^2 = 1^2 = 1$

(b)

Step1: Use given value at $x=1$

$f(1)=3$ (from piecewise definition)

(c)

Step1: Check continuity condition

A function is continuous at $x=a$ if $\lim_{x \to a} f(x) = f(a)$

Step2: Compare values

$\lim_{x \to 1} f(x)=1$, $f(1)=3$, so $1
eq 3$

Answer:

(a) $\boldsymbol{11}$
(b) $\boldsymbol{0}$
(c) $\boldsymbol{4}$

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Problem 10