Calculus help: use squeeze theorem to verify limit?

The idea of the squeeze theorem is that if you want to find the limit of a function F at the point x, then if you know two functions G and H that share the same limit L at x and G < F < H (less than or equal to), then F must have that same limit L at x. That is, F is "squeezed" between G and H.

In this case, the function we want to verify is F= x^2+1 and the hint tells us that either G=|x|+1 or H=|x|+1. Since around the point x=0, |x| is greater than x^2 (this is not true further than 1 unit away from x=0, but it only needs to be true in a small neighborhood), then we want H=|x|+1.

Now we need one more function G such that G<F on that same neighborhood and for G to have the limit L=1 at x=0. A really simple function that does that is simply the constant function G=1. So since G< F < H and lim(1)=1 (as x->0) and lim(|x|+1)=1 (as x-> 0) then lim(x^2+1)=1 (as x->0).

Sandwich Theorem Calculus Limits

lim x->0, 1<= f(x) <= g(x)

lim x->0, 1<= x^2+1 <= |x| + 1

since x^2 <= |x| for all |x|<1

The left edge and the staggering edge could be equivalent and opposite sign... whether that's no longer the case in this exercising. yet all of us comprehend that the decrease of x->0 (a million-cosx)/x^2 is a million/2 steps: we will replace the numerator: a million - cos x = 2 [sin(x/2)]^2 Lim (a million-cosx)/x^2 = Lim {2 [sin(x/2)]^2}/x^2 Lim {2 [sin(x/2)]^2}/x^2 = Lim [sin(x/2)]^2/(x^2/2) Lim [sin(x/2)]^2/(x^2/2) = lim[sin(x/2)/(x/2)]*lim[sin(x/2)/2*(x/2)... with the aid of fact the elementary decrease is: lim [sin f(x)]/f(x) = a million, if f(x) -> 0, we will get:a million/2 So i anticipate that the decrease of a million/2 - (x^2/24) is a million/2 with the aid of fact x->+0 and -0 0.000000001/24 and -0.00000000001/24 is 0!!! So the decrease is 0.5 so lim x->0 (a million/2)-(x^2/24)<Lim (a million-cosx)/(x^2)<Lim(a million/2) And its kinda obvious on the staggering edge that the lim of a million/2 is a million/2 which potential the left edge could be equivalent to the staggering.... reliable success!!!

| (x) , if x≥ 0

|x| |

| -(x), if x≤ 0

lim -(x) +1 ≤ (x^2 + 1)≤ lim (x)+1

x->0..............................x-->0

=1..................................=1

therefore lim (x^2 + 1)

................x--> 0 .........=1