Equivalent condition for the existence of left hand limit. Let $f : [a,b] \longrightarrow \mathbb R$ be a function. Let $c \in (a,b].$ Then I know that $\lim\limits_{x \to c^-} f(x) = l$ iff for every sequence $\{x_n\}_{n \geq 1}$ converging to $c$ from below we have $f(x_n) \to l.$ Now my question is $:$ Instead of taking any sequence $\{x_n\}_{n \geq 1}$ converging to $c$ from below if we take any sequence $\{x_n \}_{n \geq 1}$ increases to $c$ do we have the same conclusion as above i.e. can it still be said that $\lim\limits_{x \to c^-} f(x) = l\ $?
In order to show that it is true we need to show that if for any sequence $\{x_n \}_{n \geq 1}$ increases to $c$ the sequence $\{f(x_n)\}_{n \geq 1}$ converges to $l$ then for any sequence $\{x_n\}_{n \geq 1}$ converging to $c$ from below the sequence  $\{f(x_n)\}_{n \geq 1}$ also converges to $l.$ But I don't know how to show that.
Any suggestion in this regard will be appreciated. Thanks!
 A: 
Let $\,f:[a,b]\to\mathbb R\,$ be a function and let $\,c\in(a,b]\,.$
If for any sequence $\,\{x_n\}_{n\geqslant1}\subseteq[a,b]\,$ monotonically increasing and converging to $\,c\,$ we have $\,f(x_n)\to l\,,\,$ then $\,\lim\limits_{x\to c^-}f(x)=l\,.$

Proof :
We proceed by contradiction and assume that $\,\lim\limits_{x\to c^-}f(x)\ne l\,,$ hence ,
$\exists\;\varepsilon^*>0\,:\,\forall\,\delta>0\;\;\exists\,x_\delta\in[a,b]\cap(c-\delta,c)\;$ such that $\,\left|f(x_\delta)-l\right|\geqslant\varepsilon^*.$
In particular, for $\,\delta=1>0\,,\,$ we get $\,x_1\in[a,b]\;$ such that
$c-1<x_1<c\quad$ and $\quad\left|f(x_1)-l\right|\geqslant\varepsilon^*.$
For $\,\delta=\min\left\{\frac12,c-x_1\right\}>0\,,\,$ we get $\,x_2\in[a,b]\;$ such that
$x_1<x_2\quad,\quad c-\frac12<x_2<c\quad$ and $\quad\left|f(x_2)-l\right|\geqslant\varepsilon^*.$
For $\,\delta=\min\left\{\frac13,c-x_2\right\}>0\,,\,$ we get $\,x_3\in[a,b]\;$ such that
$x_2<x_3\quad,\quad c-\frac13<x_3<c\quad$ and $\quad\left|f(x_3)-l\right|\geqslant\varepsilon^*.$
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For $\,\delta=\min\!\left\{\frac1n,c\!-\!x_{n-1}\right\}\!>\!0\,,\,$ we get $\,x_n\in[a,b]\;$ such that
$x_{n-1}<x_n\quad,\quad c-\frac1n<x_n<c\quad$ and $\quad\left|f(x_n)-l\right|\geqslant\varepsilon^*.$
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In this way we have got a sequence $\,\{x_n\}_{n\geqslant1}\subseteq[a,b]\,$
monotonically increasing such that
$c-\frac1n<x_n<c\quad$ and $\quad\left|f(x_n)-l\right|\geqslant\varepsilon^*\quad$ for any $\,n\in\mathbb N\,.$
Consequently ,
$\{x_n\}_{n\geqslant1}\,$ is a sequence monotonically increasing and converging to $\,c\,,\,$ but $\,f(x_n)\not\to l\,.$
This is a contradiction, so the proof is complete.
