Prove that $\lim_{x \to 1}(2x+3)=5$

Question

Please check my proof

There exist $\delta $ such that $0<x<\delta \rightarrow |2x+3-5|=|2x-2|<\epsilon $

$$0<x<\delta \rightarrow 2|x-1|< \epsilon $$ $$ \rightarrow |x-1|<\frac{\epsilon }{2}$$

Choose $\delta =\frac{\epsilon }{2}$

For x is real number and $$0<x<\delta \rightarrow |2x+3-5|=|5-5|=0<\delta =\frac{\epsilon }{2}$$

Then limit is 5

Answer 1

$$|f(x)-f(1)|=|2x+3-5|=2|x-1|$$

So for a given $\epsilon$ if you choose $\delta < \epsilon /2$ we will get

$$|x-1|<\delta \rightarrow |f(x)-f(1)| <\epsilon$$