# Let $g:A\rightarrow R$ and assume f is a bounded function on$A\subseteq R$. Show that if$\lim_{x \to c}g(x)=0$, then $\lim_{x \to c }g(x)f(x) =0$

Please check my proof

let $\epsilon > 0$,$\delta >0$

since f(x) is bounded function then we have

$$0<x<\delta \leftrightarrow |f(x)-L|<\sqrt{\epsilon }$$

and g has limit equal 0

$$0<x<\delta \leftrightarrow |g(x|<\sqrt{\epsilon }$$

for $\lim_{x\rightarrow c}g(x)f(x)=0 $$0< x< \delta \leftrightarrow |f(x)-L||g(x|)< \sqrt{\epsilon }\sqrt{\epsilon }=\epsilon$$ Choose$\delta =\epsilon $then$|f(x)-L||g(x|<\epsilon $or limit equal 0 • Your proof is not well organized. You don't need to assume$\delta$. Next, your definition of a bounded function is not correct. And lot more... Commented Mar 13, 2017 at 16:39 • I forgot f(x) is bounded function -*- Commented Mar 13, 2017 at 16:43 ## 2 Answers Since$f$is bounded on$A$, there exists$M>0$such that$|f(x)|\leq M$for all$x\in A$. Let$\epsilon>0$. Using the assumption that$\lim_{x\to c}g(x)=0$(applied on$\frac{\epsilon}{M}$), we can find a$\delta>0$such that whenever$x\in A$and$0<|x-c|<\delta$, we have $$|g(x)|<\frac{\epsilon}{M}.$$ Hence, if$x\in A$and$0<|x-c|<\delta\$ then $$|f(x)g(x)-0|=|f(x)g(x)|=|f(x)|\cdot|g(x)|\leq M|g(x)|<M\frac{\epsilon}{M}=\epsilon.$$ This proves that $$\lim_{x\to c}f(x)g(x)=0.$$

Solve like this Since ff is bounded on AA, there exists M>0M>0 such that |f(x)|≤M|f(x)|≤M for all x∈Ax∈A. Let ϵ>0ϵ>0. Using the assumption that limx→cg(x)=0limx→cg(x)=0 (applied on ϵMϵM), we can find a δ>0δ>0 such that whenever 0<|x−c|<δ0<|x−c|<δ, we have |g(x)|<ϵM. |g(x)|<ϵM. Hence, if 0<|x−c|<δ0<|x−c|<δ then |f(x)g(x)−0|=|f(x)g(x)|=|f(x)|⋅|g(x)|≤M|g(x)|