Let $$ f(x) = \left\{ \begin{array}{ll} x+1 & \mbox{if } x > 1 \\ x & \mbox{if } x < 1 \end{array} \right. $$ Show that $$\lim_{x \to 1} f(x) \neq \frac{3}{2}$$
My proof
I need to prove that there exists no possible $\delta>0$ and $\varepsilon>0$ combination that satisfies $$ \left|f(x)-\frac{3}{2}\right|<\varepsilon \text{ whenever } |x-1|<\delta $$
For $1<x<1+\delta$, $f(x)=1+x$ $$ \left|f(x)-\frac{3}{2}\right|=\left|x+1-\frac{3}{2}\right|=\left|x-\frac{1}{2}\right| $$ Let $x=1+\delta_m$ such that $0<\delta_m<\delta$ $$ \left|f(x)-\frac{3}{2}\right|=\left|x-\frac{1}{2}\right|=\left|1+\delta_m-\frac{1}{2}\right|=\delta_m+\frac{1}{2} \\ \implies \left|f(x)-\frac{3}{2}\right|>\frac{1}{2} $$
For $1-\delta<x<1$, $f(x)=x$ $$ \left|f(x)-\frac{3}{2}\right|=\left|x-\frac{3}{2}\right| $$ Let $x=1-\delta_m$ such that $0<\delta_m<\delta$ $$ \left|f(x)-\frac{3}{2}\right|=\left|x-\frac{3}{2}\right|=\left|1-\delta_m-\frac{3}{2}\right|=\left|-\delta_m-\frac{1}{2}\right|=\delta_m+\frac{1}{2} \\ \implies \left|f(x)-\frac{3}{2}\right|>\frac{1}{2} $$
As we can see, $\left|f(x)-\frac{3}{2}\right|>\frac{1}{2}$ always and hence we can not find $\delta>0$ for $\varepsilon \leq \frac{1}{2}$
Is this a legit proof? Or am I wrong somewhere?
Actually, what you are supposed to prove is that there is some $\varepsilon>0$ such that, for any $\delta>0$, there is some $x\in\Bbb R$ such that $\left|x-1\right|<\delta$ and $\left|f(x)-\frac32\right|\geqslant\varepsilon$. Take $\varepsilon=\frac12$. If $\delta>0$, take $x<1$. Then $f(x)<1$, and therefore $\left|f(x)-\frac32\right|>\frac12=\varepsilon$.