I'm a little confused on these specific problems... I know the process that it takes to find Delta and how to show proof, but I'm stuck with these two math problems.
- Find the limit L, then use the Epsilon-Delta definition to prove that the limit is L. I'm trying to prove that for every Epsilon>0, there exists a Delta such that if 0<|x-a|<d, then |f(x)-L|<Epsilon.
1) lim x->4 (x+2)
On this problem, I know the limit is 6 from plugging in 4 for x. So I want to find |x-4|<d, right?
- I started off with the idea:
|(x+2)-6|<Epsilon
- That gave me |x-4|<Epsilon and this is where I'm stuck. Epsilon can't just be equal to Delta, right? There's nothing to divide the |x-4| by to get a Epsilon divided by whatever or an Epsilon times whatever, which would be our Delta. I don't know where to go from here...
2) lim x->-4 (x^2+4x) - "<- the +4x is not part of the exponent"
For this problem, I don't know if the limit is aloud to be 0 or not. Plugging in a -4 for x gives us 0 for the limit. Anyways, I want to find |x+4|<d, right?
- If the limit is aloud to be 0, I'd start off with:
|(x^2+4x)-0|<Epsilon
- Next, I factored it and got |x(x+4)|<Epsilon and now I'm stuck again, because I know there can't be an x on the right side, so I can't divide out the x, right... ?
Any tips would be great, thanks!
- Find the limit L, then use the Epsilon-Delta definition to prove that the limit is L. I'm trying to prove that for every Epsilon>0, there exists a Delta such that if 0<|x-a|<d, then |f(x)-L|<Epsilon.
1) lim x->4 (x+2)
On this problem, I know the limit is 6 from plugging in 4 for x. So I want to find |x-4|<d, right?
- I started off with the idea:
|(x+2)-6|<Epsilon
- That gave me |x-4|<Epsilon and this is where I'm stuck. Epsilon can't just be equal to Delta, right? There's nothing to divide the |x-4| by to get a Epsilon divided by whatever or an Epsilon times whatever, which would be our Delta. I don't know where to go from here...
2) lim x->-4 (x^2+4x) - "<- the +4x is not part of the exponent"
For this problem, I don't know if the limit is aloud to be 0 or not. Plugging in a -4 for x gives us 0 for the limit. Anyways, I want to find |x+4|<d, right?
- If the limit is aloud to be 0, I'd start off with:
|(x^2+4x)-0|<Epsilon
- Next, I factored it and got |x(x+4)|<Epsilon and now I'm stuck again, because I know there can't be an x on the right side, so I can't divide out the x, right... ?
Any tips would be great, thanks!