if 1≤f(x)≤x^2+2x+2 for all x, find lim as x approaches 1 of f(x). the theorem is the squeeze thoerem but how do u do this?

Question

if 1≤f(x)≤x^2+2x+2 for all x, find lim as x approaches 1 of f(x).  the theorem is the squeeze thoerem but how do u do this?

anonymous · Answer

show that the limit on the right is also 1

anonymous · Answer

which it isn't so i don't think you can show anything here

anonymous · Answer

is there a typo?  maybe 
$$x^2-2x+2$$ or 
$$x^2+2x-2$$?

anonymous · Answer

x^2 + 2x + 2

anonymous · Answer

then i do not know because that limit as x goes to 1 is 5

anonymous · Answer

well the full question is what theorem allows us to assume this? which the answer was the squeeze theorem so im not sure if that has anything to do with it..

myininaya · Answer

satellite is right you cannot apply squeeze thm here
it doesn't satisfy the if part of the theorem

anonymous · Answer

hello myininaya got any ideas here because i think it is wrong

myininaya · Answer

you know you have the following inequality, right?:
$$1 \le f(x) \le x^2+2x+2$$

anonymous · Answer

yeah, that's what was given.  i hate calculus...

myininaya · Answer

and you also have x->1,right?

anonymous · Answer

as the limit yeah

anonymous · Answer

it says to find the limit as x->1 of f(x)

myininaya · Answer

you do know that in order to use the squeeze them we must have:
$$\lim_{x \rightarrow 1}1=\lim_{x \rightarrow 1}(x^2+2x+2)$$
which is not true

anonymous · Answer

yeah calculus pretty much sucks.  especially calc two

anonymous · Answer

im a senior in high school taking statistics/calculus AKA staculus...bombed this first test, doing corrections to make up some points...ill ask the teacher tomorrow about it then...

anonymous · Answer

yeah make sure there is not a typo because if it is either of the other ones i wrote then it would work

anonymous · Answer

theres probably some kind of way to solve it...i dont know...thank