Question

Can someone please answer these questions to help me understand? Please and thank you! Will mark as brainliest!!

Answer 1

QUESTION 1  

If a function is continuous at $x=a$, then $\lim_{x \to a}f(x)=f(a)$  

Let us find the limit first,  

$\lim_{x \to 4} \frac{x-4}{x+5}$  

As $x \rightarrow 4$, $x-4 \rightarrow 0$,$x+5 \rightarrow 9$ and $f(x) \rightarrow \frac{0}{9}=0$  

$\therefore \lim_{x \to 4} \frac{x-4}{x+5}=0$  

Let us now find the functional value at $x=4$  

$f(4)=\frac{4-4}{4+5} =\frac{0}{9}=0$  

Since  

$\lim_{x \to 4} f(x)=\frac{x-4}{x+5}=f(4)$, the function is continuous at $a=4$.  

QUESTION 2  

The correct answer is table 2. See attachment.

In this table the values of x approaches zero from both sides.

This can help us determine if the one sided limits are approaching the same value.

As we are getting closer and closer to zero from both sides, the function is approaching 2.

The values are also very close to zero unlike those in table 4.

The correct answer is B

QUESTION 3

We want to evaluate;

$\lim_{x \to 1} \frac{x^3+5x^2+3x-9}{x-1}$

using the properties of limits.

A direct evaluation gives $\frac{1^3+5(1)^2+3(1)-9}{1-1}=\frac{0}{0}$.

This indeterminate form suggests that, we simplify the function first.

We factor to obtain,

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

We cancel common factors to get,

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

$=(1+3)^2=16$

The correct answer is D

QUESTION 4

We can see from the table that as x approaches $-2$ from both sides, the function approaches $-4$

Hence the limit is $-4$.

See attachment

The correct answer is option A