Show that the limit of f is equal to y

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In summary, the conversation discusses a proof for the convergence of a sequence of functions. One solution is provided by an assistant professor and another one is proposed by the person asking the question. Both solutions use the concept of uniform convergence and make use of inequalities to show the desired result. The person asking the question also points out a potential error in the assistant professor's solution.
  • #1
evinda
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Hello! ;)
Suppose that $A \subseteq \mathbb{R}, f_n \to f$ uniformly in $A$, $x_0$ an accumulation point of $A$ and $ \displaystyle \lim_{x \to x_0} {f_n(x)}=y_n \in \mathbb{R}$. $(y_n)$ converges in $\mathbb{R}$.Show that $ \displaystyle \lim_{x \to x_0} {f(x)}= \lim_{n \to +\infty} {y_n} \Rightarrow \lim_{x \to x_0} \lim_{n \to +\infty} {f_n(x)}= \lim_{n \to +\infty} \lim_{x \to x_0} f_n(x)$.

That is the solution that the assistant of the professor gave us:
  • We set $ \displaystyle y= \lim_{n \to + \infty} y_n$

    We have $|f(x)-y|=|f(x)-y_n+y_n-y| \leq |f(x)-y_n|+|y_n-y|$

    $ \displaystyle \Rightarrow \lim_{x \to x_0} \sup_{x}{|f(x)-y|} \leq \lim_{x \to x_0} \sup_{x} |f(x)-y_n|+ \lim_{x \to x_0} \sup_{x} |y_n-y| \overset{n \to \infty}{\longrightarrow} 0+0=0$

    We have used that $ \displaystyle \lim_{x \to x_0} \sup_{x} {|f(x)-y_n|}=0$,because:

    $|f(x)-y_n|=|f(x)-f_n(x)+f_n(x)-y_n| \leq |f(x)-f_n(x)|+|f_n(x)-y_n|$

    $ \displaystyle \lim_{ x \to x_0} \sup_{x} |f(x)-y_n| \leq \lim_{x \to x_0} \sup_{x} |f(x)-f_n(x)|+ \lim_{x \to x_0} \sup_{x} |f_n(x)-y_n|$.

    But, $ \displaystyle \lim_{x \to x_0} \sup_{x} f(x)= \sup_{x} \{ \lim_{n \to + \infty}{f(z_n)} | z_n \to x_0 \}$

    Therefore, we have $ \displaystyle \lim_{x \to x_0} \sup_{x} |f(x)-y_n| \leq \sup_{x} |f(x)-f_n(x)|\overset{n \to \infty}{\longrightarrow} 0 $Is this right? I am not really sure... I would do it like that:
  • We set $ \displaystyle y= \lim_{n \to + \infty} y_n$

    We want to show : $ \displaystyle \lim_{x \to x_0} f(x)= \lim_{n \to +\infty} y_n \Rightarrow \lim_{x \to x_0}f(x)=y$

    $|f(x)-y|=|f(x)-y_n+y_n-y| \leq |f(x)-y_n|+|y_n-y| (1)$

    Let $\epsilon>0$. As $y_n \to y, \exists n_0$ such that $ \forall n \geq n_0: |y_n-y|< \epsilon$

    $ \displaystyle |f(x)-y_n|=|f(x)-f_n(x)+f_n(x)-y_n| \leq |f_n(x)-f(x)|+|f_n(x)-y_n|$

    $ \displaystyle \leq \sup_{x} |f_n(x)-f(x)| + |f_n(x)-y_n| \leq \epsilon $, because:

    $f_n \to f$ uniformly in $A$, so $ \displaystyle \sup_{x} |f_n(x)-f(x)| \leq \epsilon$.
    Also, $ \displaystyle \lim_{x \to x_0} {f_n(x)}=y_n \in \mathbb{R}$, so $|f_n(x)-y_n|< \epsilon$.

    From the relation $(1)$,we have $|f(x)-y| \leq \epsilon$.

Could you tell me if it is right?
 
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  • #2
evinda said:
Hello! ;)
Suppose that $A \subseteq \mathbb{R}, f_n \to f$ uniformly in $A$, $x_0$ an accumulation point of $A$ and $ \displaystyle \lim_{x \to x_0} {f_n(x)}=y_n \in \mathbb{R}$. $(y_n)$ converges in $\mathbb{R}$.Show that $ \displaystyle \lim_{x \to x_0} {f(x)}= \lim_{n \to +\infty} {y_n} \Rightarrow \lim_{x \to x_0} \lim_{n \to +\infty} {f_n(x)}= \lim_{n \to +\infty} \lim_{x \to x_0} f_n(x)$.

That is the solution that the assistant of the professor gave us:
  • We set $ \displaystyle y= \lim_{n \to + \infty} y_n$

    We have $|f(x)-y|=|f(x)-y_n+y_n-y| \leq |f(x)-y_n|+|y_n-y|$

    $ \displaystyle \Rightarrow \lim_{x \to x_0} \sup_{x}{|f(x)-y|} \leq \lim_{x \to x_0} \sup_{x} |f(x)-y_n|+ \lim_{x \to x_0} \sup_{x} |y_n-y| \overset{n \to \infty}{\longrightarrow} 0+0=0$

    We have used that $ \displaystyle \lim_{x \to x_0} \sup_{x} {|f(x)-y_n|}=0$,because:

    $|f(x)-y_n|=|f(x)-f_n(x)+f_n(x)-y_n| \leq |f(x)-f_n(x)|+|f_n(x)-y_n|$

    $ \displaystyle \lim_{ x \to x_0} \sup_{x} |f(x)-y_n| \leq \lim_{x \to x_0} \sup_{x} |f(x)-f_n(x)|+ \lim_{x \to x_0} \sup_{x} |f_n(x)-y_n|$.

    But, $\color{red}{ \displaystyle \lim_{x \to x_0} \sup_{x} f(x)= \sup_{x} \{ \lim_{n \to + \infty}{f(z_n)} | z_n \to x_0 \}}$

    Therefore, we have $ \displaystyle \lim_{x \to x_0} \sup_{x} |f(x)-y_n| \leq \sup_{x} |f(x)-f_n(x)|\overset{n \to \infty}{\longrightarrow} 0 $Is this right? I am not really sure... I would do it like that:
  • We set $ \displaystyle y= \lim_{n \to + \infty} y_n$

    We want to show : $ \displaystyle \lim_{x \to x_0} f(x)= \lim_{n \to +\infty} y_n \Rightarrow \lim_{x \to x_0}f(x)=y$

    $|f(x)-y|=|f(x)-y_n+y_n-y| \leq |f(x)-y_n|+|y_n-y| (1)$

    Let $\epsilon>0$. As $y_n \to y, \exists n_0$ such that $ \forall n \geq n_0: |y_n-y|< \epsilon$

    $ \displaystyle |f(x)-y_n|=|f(x)-f_n(x)+f_n(x)-y_n| \leq |f_n(x)-f(x)|+|f_n(x)-y_n|$

    $ \displaystyle \leq \sup_{x} |f_n(x)-f(x)| + |f_n(x)-y_n| \leq \epsilon $, because:

    $f_n \to f$ uniformly in $A$, so $ \displaystyle \sup_{x} |f_n(x)-f(x)| \leq \epsilon$.
    Also, $ \displaystyle \lim_{x \to x_0} {f_n(x)}=y_n \in \mathbb{R}$, so $|f_n(x)-y_n|< \epsilon$.

    From the relation $(1)$,we have $|f(x)-y| \leq \epsilon$.

Could you tell me if it is right?
Your proof looks fine (except that the very last $ \epsilon$ looks as though it should be $2 \epsilon$). The other proof goes along the same lines apart from the equation that I have coloured red. I do not see what is going on there, and I don't trust it. In any case, it does not make clear where the uniform convergence $f_n\to f$ is used, unlike your solution which pinpoints that essential component of the proof.
 
  • #3
Opalg said:
Your proof looks fine (except that the very last $ \epsilon$ looks as though it should be $2 \epsilon$). The other proof goes along the same lines apart from the equation that I have coloured red. I do not see what is going on there, and I don't trust it. In any case, it does not make clear where the uniform convergence $f_n\to f$ is used, unlike your solution which pinpoints that essential component of the proof.

Great!Thank you very much! :)
 

FAQ: Show that the limit of f is equal to y

What does it mean to show that the limit of f is equal to y?

Showing that the limit of f is equal to y means proving that as the input values of the function f approach a certain value, the output values of the function also approach a specific value, which is y. This is also known as the limit of a function and is often denoted as lim f(x) = y.

How is the limit of a function calculated?

The limit of a function can be calculated by evaluating the function at values closer and closer to the given input value. If the output values approach a specific value (y), then the limit of the function is equal to that value.

Can the limit of a function be undefined?

Yes, the limit of a function can be undefined if the output values do not approach a specific value as the input values get closer and closer. This usually occurs when there is a discontinuity or a jump in the function.

Why is it important to show the limit of a function?

The limit of a function is important because it helps determine the behavior of the function near a certain input value. It can also help in finding the maximum or minimum values of a function and determining if the function is continuous at a certain point.

What are some common techniques used to show the limit of a function?

Some common techniques used to show the limit of a function include direct substitution, factoring and simplifying, and using algebraic and trigonometric identities. In some cases, L'Hopital's rule or the squeeze theorem may also be used.

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