Continuity: Is it possible to define f(x)=sin(3x^2-3)/x^201 at x=0 so f cont. at x=1?

[confusedcalculusstudent]

I'm having trouble with this problem:

Is it possible to define f(0) in such a way that would make f(x)=sin(3x^2-3)/x^201 continuous at x=1? If so, what should f(0) be so that f(x) is continuous at x=1?

For starters, I thought that saying f(0) meant that x=0? (please correct me, I'm in an intro math class for a reason XD)

Assuming x=1, I plugged it in and got: sin(3(1)^2-3)/(1)^2-1 => sin(0)/0, which doesn't seem in line with what they're asking for
Assuming x=0, sin(3(0)^2-3)/(0)^2-1 => sin(-3)/-1... multiplied by 3/3, 3sin(-3)/-3, (can sinh/h=1 be invoked here? I tried that and got an answer of 3, but I don't know if that's correct or how to justify that with the question saying x=1)

I'm very confused, please help

 

[lev888]

Yes, f(0) is the value of f(x) when x = 0.

 

[Dr.Peterson]

I'm having trouble with this problem:

Is it possible to define f(0) in such a way that would make f(x)=sin(3x^2-3)/x^201 continuous at x=1? If so, what should f(0) be so that f(x) is continuous at x=1?

For starters, I thought that saying f(0) meant that x=0? (please correct me, I'm in an intro math class for a reason XD)

Assuming x=1, I plugged it in and got: sin(3(1)^2-3)/(1)^2-1 => sin(0)/0, which doesn't seem in line with what they're asking for
Assuming x=0, sin(3(0)^2-3)/(0)^2-1 => sin(-3)/-1... multiplied by 3/3, 3sin(-3)/-3, (can sinh/h=1 be invoked here? I tried that and got an answer of 3, but I don't know if that's correct or how to justify that with the question saying x=1)

I'm very confused, please help

I think you meant f(x)=sin(3x^2-3)/(x^2-1). Right? Always check for typos, and use parentheses as needed!

Since sin(0)/0 = 0/0 is undefined, the function as given is not defined at x=0. As you suggested (but stated incorrectly), you need to use the fact that the limit of sin(h)/h = 0 as h approaches 0. You didn't carry out that limit (which must be as x approaches 1, not 0), but if you can show that the limit is 3, this does allow you to conclude that \(\displaystyle \displaystyle \lim_{x\rightarrow 1} f(x) = 3\), so that in order to make f continuous at x=1, f(1) must be defined as 3.

It is not true that sin(h)/h = 1, so you can't say that 3sin(-3)/-3 = 3.

Do you understand what continuity means? Do you understand how to use the fact you referred to to find the limit of f?

 

[Steven G]

I'm having trouble with this problem:

Is it possible to define f(0) in such a way that would make f(x)=sin(3x^2-3)/x^201 continuous at x=1? If so, what should f(0) be so that f(x) is continuous at x=1?

For starters, I thought that saying f(0) meant that x=0? (please correct me, I'm in an intro math class for a reason XD)

Assuming x=1, I plugged it in and got: sin(3(1)^2-3)/(1)^2-1 => sin(0)/0, which doesn't seem in line with what they're asking for
Assuming x=0, sin(3(0)^2-3)/(0)^2-1 => sin(-3)/-1... multiplied by 3/3, 3sin(-3)/-3, (can sinh/h=1 be invoked here? I tried that and got an answer of 3, but I don't know if that's correct or how to justify that with the question saying x=1)

I'm very confused, please help

I am confused as well! Yes, f(0) means x=0. Now here is why I am confused. For x to continuous at x=1, it does not matter what is happening at x=1. I think that is why you asked if f(0) means x=0? Are you sure that the problem is not Is it possible to define f(1) in such a way that would make f(x)=sin(3x^2-3)/x^201 continuous at x=1? If so, what should f(1) be so that f(x) is continuous at x=1? If the problem was stated correctly then the answer is NO. You can NOT define f(0) to make f(x) continuous at x=1.

 

[Steven G]

I think you meant f(x)=sin(3x^2-3)/(x^2-1). Right? Always check for typos, and use parentheses as needed!

Since sin(0)/0 = 0/0 is undefined, the function as given is not defined at x=0. As you suggested (but stated incorrectly), you need to use the fact that the limit of sin(h)/h = 0 as h approaches 0. You didn't carry out that limit (which must be as x approaches 1, not 0), but if you can show that the limit is 3, this does allow you to conclude that \(\displaystyle \displaystyle \lim_{x\rightarrow 1} f(x) = 3\), so that in order to make f continuous at x=1, f(1) must be defined as 3.

It is not true that sin(h)/h = 1, so you can't say that 3sin(-3)/-3 = 3.

Do you understand what continuity means? Do you understand how to use the fact you referred to to find the limit of f?

Can you please explain why what is happening at x=0 will make f(x) continuous at x=1???