Finding the Limit of (2x + 1)/(x + 4) at x = -4?

[nycmathdad]

Find the limit of (2x + 1)/(x + 4) as x tends to - 4 from the right side.

I know there's a vertical asymptote at x = -4. I think the best way to solve this problem is by graphing the function. I am not too sure about how to solve algebraically.

I am thinking about the number line.

<---------------(-4)----------------->

What if I select values to the left and right of -4 but not including -4? By doing this, I will then know if the interval (-00, -4) is positive or negative and if the interval (-4, 00) is positive or negative.

Is this correct so far?

 

[jonah1]

Beer soaked fill in the boxes hint follows.

Find the limit of (2x + 1)/(x + 4) as x tends to - 4 from the right side.

I know there's a vertical asymptote at x = -4. I think the best way to solve this problem is by graphing the function. I am not too sure about how to solve algebraically.

I am thinking about the number line.

<---------------(-4)----------------->

What if I select values to the left and right of -4 but not including -4? By doing this, I will then know if the interval (-00, -4) is positive or negative and if the interval (-4, 00) is positive or negative.

Is this correct so far?

As x approaches −4 from the $\boxed{?}$, 2x + 1 approaches $\boxed{??}$ and x + 4 approaches $\boxed{?}$ from the $\boxed{?}$.
Therefore, the ratio $\frac{2x+1}{x+4}$ becomes $\boxed{?}$ in the $\boxed{?}$ $\boxed{?}$, so
$\mathop {\lim }\limits_{x \to -4^+ } \frac{2x+1}{{x+4 }} = \boxed{??}$

Note: Number of question marks corresponds to number of letters or symbols.

To support whatever conclusion you get from filling in the boxes, make a table with the following values of x: -3.9, -3.99, -3.999

 

[HOI]

"Approaching from the right side" means that x is slightly larger than -4. Write x= -4+ d with d> 0. Then $\frac{2x+ 1}{x+ 4}= \frac{2(-4+ d)+ 1}{-4+ d+ 4}= \frac{-8+ 2d}{d}= \frac{-8}{d}+ 2$.

x going to -4 means d going to 0. As d goes to 0, $\frac{-8}{d}$ goes to NEGATIVE infinity. Of course, the "+ 2" is irrelevant.

 

[nycmathdad]

"Approaching from the right side" means that x is slightly larger than -4. Write x= -4+ d with d> 0. Then $\frac{2x+ 1}{x+ 4}= \frac{2(-4+ d)+ 1}{-4+ d+ 4}= \frac{-8+ 2d}{d}= \frac{-8}{d}+ 2$.

x going to -4 means d going to 0. As d goes to 0, $\frac{-8}{d}$ goes to NEGATIVE infinity. Of course, the "+ 2" is irrelevant.

What does d mean in your reply?

 

[HOI]

"Write x= -4+ d". d is how much larger x is than -4. For example, if x= -3.75, d= 0.25; if x= -3.9, d=0.1; if x= -3.95, d= 0.05; if x= -3.99, d= 0.01. As x goes to -4, d goes to 0.

 

[nycmathdad]

"Write x= -4+ d". d is how much larger x is than -4. For example, if x= -3.75, d= 0.25; if x= -3.9, d=0.1; if x= -3.95, d= 0.05; if x= -3.99, d= 0.01. As x goes to -4, d goes to 0.

Can you please look at my Limit of Rational Function parts 1 to 5 and tell if any of that makes sense. It's important for me to fully grasp Section 1.5 before moving on to Section 1.6 which involves the fuzzy epsilon-delta definition of a limit.