I got the right answer but confused about logic

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kory

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If I'm approaching 1 from the left, then that means I should be able to plug 0 into the function. So why does 0 not work but 1 does?
 

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The limit asked me to approach from the left of 1. Left of 1 is zero....
 
It also states that x can be <= to 1. Therefore any number less than 1 should be a number that I could plug in. So I chose 0 and it failed. If thats the case then the limit is NOT a number <= to 1. Its just 1.
 
kory said:
The limit asked me to approach from the left of 1. Left of 1 is zero....
Click to expand...
kory said:
It also states that x can be <= to 1. Therefore any number less than 1 should be a number that I could plug in. So I chose 0 and it failed. If thats the case then the limit is NOT a number <= to 1. Its just 1.
Click to expand...

In calculus, "number" does not mean integer! To the left of 1 are numbers like 0.9999999999999. As x approaches 1 from the left, getting closer and closer through numbers like that, what does the function approach? What will happen if you just plug in 1?

You might try graphing the function.
 
kory said:
It also states that x can be <= to 1. Therefore any number less than 1 should be a number that I could plug in. So I chose 0 and it failed. If thats the case then the limit is NOT a number <= to 1. Its just 1.
Click to expand...
The notation \(\mathop {\lim }\limits_{x \to \bf{1^ - }} \) means that \(x\) is close to \(1\) on the left (the negative) side.
It may not be clear to you but \(0\) is not close to \(1\); in fact \(0.999999\) is not close to \(1\) in mathematical terms.
What is look at the problem. If you to evaluate \(f\left(0.999999\right)\) what would you use.
I hope your answer would be \(9x+5\) or \(9(0.999999)+5\sim 14\).
You must learn that limits are never, never solved by substitution. They are solved by logical approximations.
 
Hmmm...I guess I gotta keep tinkering with it.
I got all of the answers right. Just a little confused on the wording.
 
kory said:
Hmmm...I guess I gotta keep tinkering with it.
I got all of the answers right. Just a little confused on the wording.
Click to expand...
Presumably you are learning about what limits mean, and specifically what "approach from the left" means; that's what you need to figure out.

Have you tried discovering what it means by doing the things we've suggested? One is to draw the graph, and see what happens to y as you move a point along it toward x=1. Another is to calculate values of y for values of x like 0, .5, .9, .99, ... approaching 1 from the left.

In both cases, the limit they are asking for is the number that y approaches as you do these things; what you'll discover is that, because this function is continuous to the left of x=1, that limit is exactly what you get when you plug in x=1.

What you need to see is why that is true, and how you can know it for yourself without having to ask.

What part of the wording still confuses you?
 
Dr.Peterson said:
Presumably you are learning about what limits mean, and specifically what "approach from the left" means; that's what you need to figure out.

Have you tried discovering what it means by doing the things we've suggested? One is to draw the graph, and see what happens to y as you move a point along it toward x=1. Another is to calculate values of y for values of x like 0, .5, .9, .99, ... approaching 1 from the left.

In both cases, the limit they are asking for is the number that y approaches as you do these things; what you'll discover is that, because this function is continuous to the left of x=1, that limit is exactly what you get when you plug in x=1.

What you need to see is why that is true, and how you can know it for yourself without having to ask.

What part of the wording still confuses you?
Click to expand...
Sounds like your talking about Epsilon and Delta. As a point gets approached on x then the other point moves along y.... Correct?
 
kory said:
Sounds like your talking about Epsilon and Delta. As a point gets approached on x then the other point moves along y.... Correct?
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I wasn't explicitly referring to epsilon and delta, just to the informal idea of a limit. And I want you mostly to think informally. Again, have you done any of the things we suggested, which are meant to help you do that?

But, yes, as you change x, y = f(x) also changes, so the point (x,y) moves along the graph.

The question is, do you see yet why in your problem, as x approaches 1 from the left, y will approach f(1) = 14?

And then, do you see what this is not true as you approach from the right?

Here's the graph:
1622938015203.png
 
Dr.Peterson said:
I wasn't explicitly referring to epsilon and delta, just to the informal idea of a limit. And I want you mostly to think informally. Again, have you done any of the things we suggested, which are meant to help you do that?

But, yes, as you change x, y = f(x) also changes, so the point (x,y) moves along the graph.

The question is, do you see yet why in your problem, as x approaches 1 from the left, y will approach f(1) = 14?

And then, do you see what this is not true as you approach from the right?

Here's the graph:
View attachment 27648
Click to expand...
Im taking calculus 1 class over the summer. Im only 4 days in and we move extremely fast. Some of the things you say havn't completely clicked yet..I'm getting there though...
 
Pick any point on the graph where the x value is to the left of 1. What is that y value? Then following the curve move to the right so you are getting closer to x=1 (but still less than 1). What is the y value now? Move along the curve closer to x=1, but still to the left of x=1. What is the y value. Now you are probably very close to x=1. Still move along the curve getting closer to x=1, but still to the left. What is the x-value? Now you should be extremly close to x=1, but to the left of x=1. What is the y value? What are these y values approaching? This is your (left hand) limit.
 
Please check out the video here. It was made for this particular problem. I think that it was important that you understand this concept so that you can move on with your limits.
 
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