Problem proving if a limit exists

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Homework Statement


We are given two functions f : \mathbb{R}^n -> \mathbb{R} and g: \mathbb{R}^n -> \mathbb{R}. For every x \in \mathbb{R}^n we define the following:

k(x) = max{f(x), g(x)}
h(x) = min{f(x), g(x)}

The question is:
if lim x-> a k(x) exists and lim x-> a h(x) exists, and the limits are equal, does that imply that lim x->a f(x) exists?


Homework Equations





The Attempt at a Solution


Suppose we define f(x) = 1/x and g(x) = x. At x = 0 f(x) is not defined, so g(x) is the minimum and the maximum at x = 0. Therefore k(x) = g(x) and h(x) = g(x). We know that lim x-> 0 h(x) exists and lim x-> 0 k(x) exists, but lim x -> 0 f(x) does not exists.

Is my work here correct or am i wrong in assuming that if f(0) is not defined then g(0) is the maximum and the minimum at x = 0?
 
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Tomath said:

Homework Statement


We are given two functions f : \mathbb{R}^n -> \mathbb{R} and g: \mathbb{R}^n -> \mathbb{R}. For every x \in \mathbb{R}^n we define the following:

k(x) = max{f(x), g(x)}
h(x) = min{f(x), g(x)}

The question is:
if lim x-> a k(x) exists and lim x-> a h(x) exists, and the limits are equal, does that imply that lim x->a f(x) exists?

Homework Equations



The Attempt at a Solution


Suppose we define f(x) = 1/x and g(x) = x. At x = 0 f(x) is not defined, so g(x) is the minimum and the maximum at x = 0. Therefore k(x) = g(x) and h(x) = g(x). We know that lim x-> 0 h(x) exists and lim x-> 0 k(x) exists, but lim x -> 0 f(x) does not exists.

Is my work here correct or am i wrong in assuming that if f(0) is not defined then g(0) is the maximum and the minimum at x = 0?
Hello Tomath. Welcome to PF !

lim x→0 k(x) does not exist. Furthermore, this limit has nothing to do with whether or not k(0) exists, nor does it depend upon the value of k(x). From the left of x=0, k(x) approaches 0. From the right it approaches +∞ .
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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