Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

While proving that differentiability implies continuity in lecture 2, it is stated that we never use x=x0 for the proof so as to avoid the multiplying and diving by zero. But then limit x->x0 (x-x0) is replaced by zero to achieve the result. Can someone please explain this?

MIT 18.01 Single Variable Calculus (OCW)
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
I'm not sure where in the lecture this happens exactly, but I perused a few sections of it so that I could make sure I understood your use of x0, or\[x_{0}.\]When calculating a derivative using a difference quotient (d.q.), if we plug in x0, we get an indeterminate form (i.f.), which is 0/0. An i.f. does not mean the limit does not exist, it means that we need to juggle the equation around a bit. I'm not sure where the prof plugs in x0, but once the d.q. has been juggled to the point where plugging in x0 does not result in an i.f., we may plug it in at that point. Consider this:\[\lim_{x \rightarrow 1}\frac{ (x-1)*(10+x) }{ x-1 }\]If we just plug in 1, we get an i.f. But if we juggle this thing we can fix that by dividing the top and bottom by (x-1). The limit = 11.
Thanks for the explanation. I understand this portion where we bring an equation to such a form. The issue is towards the very end of the lecture 2 (from 47:50 to 49:30). Can you please take a look at it? It would be very helpful. Thanks in advance.
Ah, thanks for the clarification. Let's start with a generic form of a limit that we can work from together:\[\lim_{x \rightarrow y} f(x).\]When we ask, "What is the limit as x approaches y?", we know that the function f(x) is not required to have a value at y. Here is an example: \[\lim_{x \rightarrow 1}\frac{ x^2+9x-10 }{ x-1 }.\]This is basically the function x+10=y with a missing point at x=1. We know that this line has a limit as x approaches 1 because the line is continuous everywhere except at x=1. I believe you know this, but I am laying foundation. Now, as we consider the limit as x approaches y, in my first example, we never actually compute the value of y in the equation we're taking the limit of. And for good reason, this value of y is not required to exist in the function. So what the professor is saying, what looks like 0/0 is really a value approaching 0. There are an infinite number of values occupying the space on this approach to 0. Those are the values we are considering. Numbers such as 0.000000001/0.00000000000001 for instance. When we plug in x-x0, we never actually plug in x-x. x-x0 just gets smaller and smaller and smaller, approaching zero, infinitely.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Thanks a ton for the clarification. This resolves the doubt. P.S. : do you by any chance have a pdf of the simmons book? it has gone out of print in my country and i couldn't find it anywhere on the internet. Appreciate your help. Thanks again
Glad I could help! I'm not familiar with the simmons book I'm afraid.

Not the answer you are looking for?

Search for more explanations.

Ask your own question