# Prove that Lim x->c f(x)g(x)=0 if Lim x->c f(x)=0 and |g(x)|<K

In summary, for a range of input values within δ of c, the corresponding outputs can be made within Kϵ of 0, even if g(x) is discontinuous at c. This is possible because the product of the absolute values of f(x) and g(x) equals the absolute value of the product, and when one of the values is 0, the result is also 0. This proposition holds true as long as g(x) is bounded, even if it has jumps at x=c. This is not possible with any other finite limit except 0.
Homework Statement
Suppose Lim x->c f(x)=0 and that g(x) is bounded by a number K>0 such that |g(x)|<K. Define h by h(x)=f(x)g(x). Prove that Lim x->c h(x)=0.
Relevant Equations
|x-c|<δ -> |f(x)-0|<ϵ
|g(x)|<K
|x-c|<δ -> |f(x)-0|<ϵ
|g(x)|<K
|g(x)||f(x)-0|<K*ϵ
The product of the absolute values equals the absolute value of the product.
|f(x)g(x))|=|h(x)|<K*ϵ
For a range of input values within δ of c, the corresponding outputs can be made within Kϵ of 0. What if g(x) has a jump at x=c but is bounded by K. If approaching x=c from the right g(x) approaches L1 where |L1|<K and if approaching x=c from the left g(x) approaches L2 where |L2|<K. Will the above inequality holds because f(x)=0 and g(x) is bounded though discontinuous fg is again a zero function.

The proposition is true even if ##g## is discontinuous at ##c##.

It works with zero as a limit since ##0\cdot c = 0## as long as ##c## is any number. If ##c\in \{\pm \infty \}## then we cannot conclude anything, which is why we need that ##g(x)## is bounded. But even if it jumps around, multiplying by zero will result in zero, i.e. the gaps become smaller and smaller if we approach zero. It would not work with any other finite limit.

## 1. What does the notation "Lim x->c" mean?

The notation "Lim x->c" represents the limit of a function as x approaches a specific value, c. This means that we are interested in the behavior of the function as x gets closer and closer to the value c.

## 2. How do you prove that Lim x->c f(x)g(x)=0?

To prove that Lim x->c f(x)g(x)=0, we can use the Squeeze Theorem. This theorem states that if we have two functions, f(x) and g(x), such that Lim x->c f(x)=0 and |g(x)|c f(x)g(x)=0. This means that as long as g(x) is bounded by a constant and f(x) approaches 0, then the product f(x)g(x) will also approach 0.

## 3. Can you provide an example of using the Squeeze Theorem to prove Lim x->c f(x)g(x)=0?

Yes, for example, let's say we have the functions f(x)=x and g(x)=sin(x). As x approaches 0, f(x) approaches 0 and g(x) is bounded by -1 and 1. Therefore, according to the Squeeze Theorem, Lim x->0 f(x)g(x)=0.