Why is this claim about limits true I didn't understand why this claim from wikipedia is true

...More specifically, when $f$ is applied to any input sufficiently close to $p$, the output value is forced arbitrarily close to $L$.

Is not the contrary? we have formally from the definition of limits:

for every $\epsilon>0$ we have one $\delta>0$ such that
  $0<|x-p|<\delta\implies |f(x)-L| < \epsilon$.

So I think, the best claim is 

... More specifically, any output sufficiently close to $L$, the input
  value is forced arbitrarily close to $p$.

So my question is why the wikipedia claim is true? 
 A: The original claim is correct. Let's take a look at an example $f(x)=\sin x, p=0, L=0$. If you take any $\varepsilon$ and take your $x$ really close to $p$ ($|x|<\varepsilon$ is enough, but it's not important), then you will get that $\sin x$ is really close to $L$.
But your modified claim is not correct. We can be as close to $L$ as we like without $x$ being close to $p$. For example, we can have $x=2\pi$. Then $\sin x$ is in fact equal to $L$, but $x$ is not close to $p$.
A: Definition of limits, it words:
"If $x$ is close to $p$, then $f$ is close to $L$." 
Now, $x$ is the input to the function, so I'll replace occurrences of $x$ with those words:
"If the input to the function is close to $p$, then $f$ is close to $L$." 
Now, $f$ is the output of the function, so I'll replace those words:
"If the input to the function is close to $p$, then the output of the function is close to $L$." 
The Wikipedia sentence is not that great, but it is basically correct.
A: 
for every $\varepsilon>0$ we have one $\delta>0$ such that
  $0<|x-p|<\delta\implies |f(x)-L| < \varepsilon$.

The phrase $\text{“ } 0<|x-p|<\delta \Longrightarrow |f(x)-L|<\varepsilon \text{ ''}  $ says if $x$ is close to $p$ then $f(x)$ is close to $L$. 
The phrase $\text{“ For every } \varepsilon>0, \text{ there exists } \delta>0\ \cdots\cdots \text{ ''}$ starts with $\varepsilon$, which measures distances in the space where $f(x)$ and $L$ live, and goes on to $\delta$, which measures distances in the space where $x$ and $p$ live, and I suspect that is the cause of your confusion.
What it's saying is that no matter how close you want to make $f(x)$ to $L$ (i.e. no matter how small $\varepsilon$ is), you can do it by making $x$ close enough to $p$ (i.e. by making $\delta$ small enough).  Thus making $x$ close to $p$ makes $f(x)$ close to $L$.
With $f(x)=x^2$, you have $\lim\limits_{x\to3} f(x)=9$.  But making $f(x)$ close to $9$ is no guarantee that $x$ is close to $3$, since $x$ could be close to $-3$.
