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

[DumpmeAdrenaline]

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.

 

[PeroK]

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

 

[fresh_42]

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.