Pointwise limits in Real Analysis 4 In the following two questions I need to find the pointwise limit of the sequence defined for $x\in R$. I have had some success with the first example and believe I have done it correct. But in the second example I am not sure how to find the pointwise limit or if it is uniformly convergent
1) $f_n(x) = \dfrac{nx+x^2}{n^2}$. Then
$$ \lim\limits_{n\to\infty} f_n(x) =\lim\limits_{n\to\infty} \left(\dfrac{x}{n} + \dfrac{x^2}{n^2}\right) = x\left(\lim\limits_{n\to\infty} \dfrac{1}{n}\right) + x^2\left(\lim\limits_{n\to\infty} \dfrac{1}{n^2}\right)=0+0=0$$
So this converge pointwise to $0$.
However how do I find the pointwise limit of the following sequence and whether or not that it uniformly converges?
2) $f_n(x) = e^{-(x-n)^2}$
 A: Try fixing $x_0\in \mathbb R$ and study $\lim_{n\to +\infty} f_n(x_0)$. In the case you are asking:
$$\lim_{n\to +\infty} e^{-(x_0-n)^2} = \lim_{n\to +\infty} \frac{1}{e^{(x_0-n)^2}}=0$$
because as $n\to +\infty$, as $x_0$ is fixed, $(x_0-n)^2$ becomes larger, and so does $e^{(x_0-n)^2}$.
If you do not see it clear, you can choose some particular values of $x_0$.

So $f_n$ converges pointwise to $0$. Now, to study uniform convergence, note that the only possible limit is $0$, because uniform convergence implies pointwise convergence. So you will have to prove that
$$\sup_{x\in \mathbb R} |f_n(x)-0|=\sup_{x\in \mathbb R} |f_n(x)|\to 0$$
as $n\to \infty$.
But, in this case, $f_n(n)=1$, so $\sup_{x\in \mathbb R} |f_n(x)|>1$.
In the general case, it is usually a good idea to find $\sup |f_n(x)|$ using derivatives.
A: For 1), uniform convergence would imply that given some $\epsilon > 0$, you can find $n$ large enough so that $|f_n(x)| < \epsilon$ for all $x$. Why is this impossible? (Consider large enough $x$.)
For 2), as $n \to \infty$ the exponent tends to $-\infty$ so the pointwise limit is $0$. For uniform convergence, follow the same approach as 1). For $0<\epsilon<1$, why can't you find $n$ large enough so that $|f_n(x)| < \epsilon$ for all $x$? (See copper.hat's hint: note that $|f_n(x)| \le |f_n(n)|=1$ for all $x$ and $n$.)
