anonymous
  • anonymous
The limit of (2+x)/(1-x) as x approaches 1 from the right? Help please?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
beginnersmind
  • beginnersmind
Try sketching it. And think about what happens to the fraction as x gets close to 1. Use a calculator to calculte some values, like at x = 1.1 or x = 1.01.
anonymous
  • anonymous
I have to do it on paper without my calculator or graphing paper. Simply through math. .-.
beginnersmind
  • beginnersmind
Ok, but it still helps to know what the answer is "before" doing the math.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

beginnersmind
  • beginnersmind
You go from "How do I this?" to "I know what the answer is, now how do I justify it?" The second is much easier.
anonymous
  • anonymous
I know it's negative infinity. I just don't understand how the answer is that.
beginnersmind
  • beginnersmind
Ok, cool. Why is it negative infinity. What happens to the numerator and the denominator near 1?
anonymous
  • anonymous
They decrease?
beginnersmind
  • beginnersmind
What's lim(x+2) and lim(1-x) at x=1
anonymous
  • anonymous
3 and 0
beginnersmind
  • beginnersmind
Right. So the limit is 3/0. That's either + or - infinity, depending on whether the denominator goes to zero through positive or negative numbers.
anonymous
  • anonymous
but 3//0 is undefined. The graph has a hole at 1 and becomes asymptotic on the y-axis.
e.mccormick
  • e.mccormick
Another way to think about it logically is to start similarly to how beginners initially said with 1.1 and 1.01, but with easier numbers. You are coming in from the right, so numbers larger than 1. Say 3. (2+3)/(1-3) = -5/2. Then 2, so -4. Then look at it logically. You have negative numbers where the numerator is getting what and denominator is getting what? Larger? Smaller? Is it going to change from negative to positive and if so, when?
beginnersmind
  • beginnersmind
"but 3//0 is undefined. The graph has a hole at 1 and becomes asymptotic on the y-axis." Indeed, 3/0 is undefined for numbers. But for limits it makes sense. The function isn't defined at x = 1. But it is defined for 1.1, 1.01, 1.001, etc. When we say that the limit is minus infinity we're just saying that the value of the expression is arbitrarily small, as long as we are close enough to x=1. Another way to say that is that the graph has an asymptote at x=1.
anonymous
  • anonymous
Okie. Thank you. cx
beginnersmind
  • beginnersmind
No problem. Hope I managed to make it a little clearer at least.
e.mccormick
  • e.mccormick
A limit is not where you are at but where you are heading. That is where my logically comment comes in. Again, to tie it to the very good explanation by beginners, this line is travelling down the x numbers pretty rapidly from say 100 to about 4, but the y numbers travele slowly. From x=4 to x=2, the x and y change at about the same speed. Below x=2, the x is changing very slow in comparison to how fast the y is changing. Then, it venetially hits x=1 where it is, as you pointed out perfectly, undefined. So the limit is headding to something undefined. In this case, that particular undefined is negative infinity. That is why it makes sense for a limit. You are heading to the undefined value of negative infinity. You can't define it close than that.

Looking for something else?

Not the answer you are looking for? Search for more explanations.