# Continuity and Differentiability

RandomGuy1

## Homework Statement

f(x) = sin ∏x/(x - 1) + a for x ≤ 1
f(x) = 2∏ for x = 1
f(x) = 1 + cos ∏x/∏(1 - x)2 for x>1

is continuous at x = 1. Find a and b

## Homework Equations

For a lim x→0 sinx/x = 1.

## The Attempt at a Solution

I tried equating the two limits, x →1+ and x→1- to each other and to 2∏. Didn't work.
Maybe I something wrong, don't have any clue on how to proceed now. Can I get a clue?

## Answers and Replies

Homework Helper
First let's get the notation straight. By sin ∏x/(x - 1) + a, do you mean $$\frac{\sin(\pi x)}{x - 1} + a?$$
And then 1 + cos ∏x/∏(1 - x)2, is that $$1 + \frac{\cos(\pi x)}{\pi(1 - x)^2}$$ or $$1 + \frac{\cos(\pi x)}{\pi} (1 - x)^2?$$

Anyway, you should be able to find the limit as ##x \to 1## for both parts separately. For example, try showing that
$$\lim_{x \to 1} \frac{\sin \pi x}{x - 1} = - \pi.$$

As you have already correctly remarked, continuity of f implies that
$$\lim_{x \to 1} f(x) = f(1) = 2 \pi$$
so you should then choose a and b such that these values are equal to ##2\pi##.

• 1 person
RandomGuy1
Oops, sorry, it was [1 + cos∏x]/{[∏(1 - x)2}

I put x = (1 - h) and as you said got lim x → 1- = a -∏
and lim x → 1+ = ∏/2 + b.

Thus, I got a = 3∏ and b = 3∏/2, which happen to be the right answer. Thank you!