Uniform limits do not necessarily preserve differentiability, that is, even though $f_{n} \rightrightarrows f$, we may not have immediately that $f_{n}'\rightrightarrows f'$ or even $f'_{n} \to f'$ (take for example $f_{n}(x) = \sqrt{ x^{2}+\frac{1}{n} }$ or $f_{n}(x)={\sin(n^2x)}/{x}$). > [!NOTE] Statement 1 (For $(f_{n})$ in $C^1,$ $f_{n}\to f$ and $f'_{n} \rightrightarrows g$ implies $f'=g$) > Let $(f_{n})$ be a [[Sequences|sequence]] in $C^1([a,b])$ (the set of [[Continuous Differentiability|continuously differentiable real functions]], understood as a one-sided derivative). Suppose $(f_{n})$ [[Pointwise Convergence of Sequence of Real Functions doesn't Imply Uniform Convergence|converges pointwise]] to $f$ and that $f_{n}'$ [[Uniform Convergence of Sequence of Real Functions|converges uniformly]] to $g.$ Then $f'=g$ (or equivalently, $f_{n}'\to f'$) and so $f$ is $C^1.$ ###### Proof Since $f_n^{\prime} \rightrightarrows g$, [[Uniform Convergence preserves Riemann Integrability|this theorem]] yields $\int_a^x g=\int_a^x \lim _{n \rightarrow \infty} f_n^{\prime}=\lim _{n \rightarrow \infty} \int_a^x f_n ',$which by the [[Fundamental theorem of calculus|second FTC]] yields $\int_a^x g=\lim _{n \rightarrow \infty}\left[f_n(x)-f_n(a)\right]=f(x)-f(a).$Since $g$ is continuous ($f_{n}'$ is continuous so it follows from [[Uniform Limit of Sequence of Continuous Real Function is Continuous|here]] that $g$ is continuous which yields that $f$ is also continuous.), applying the [[Fundamental theorem of calculus|first FTC]] (differentiating both sides) yields that $f$ is differentiable and that $g(x)=f'(x).$ You can also avoid integrals entirely: ###### Proof 1. Fix $x$ and $h\neq0$.  By the Mean Value Theorem, for each n there is $\xi_n$ between $x$ and $x+h$ so that $\frac{f_n(x+h)-f_n(x)}{h} \;=\;f_n^{\prime}(\xi_n).$ 2. As $n\to\infty,$ the left side tends to $\frac{f(x+h)-f(x)}{h}$ (using pointwise convergence at $x$ and $x+h$), while the right side tends to $g(\xi_n)$.  But since $\xi_n\in[x,x+h]$ and $g$ is continuous (by uniform convergence), $g(\xi_n)\to g(x)$ as $h\to0$. 3. Thus $\lim_{h\to0}\frac{f(x+h)-f(x)}{h} \;=\; g(x),$ showing f is differentiable at x with $f^{\prime}(x)=g(x).$ # Application(s) **Corollary**: [[Uniform Convergence and Differentiabilty for Series of Real Functions]] asserts that $\left( \sum_{k=1}^{\infty}f_{k} \right)'=\sum_{k=1}^{\infty} f_{k}'$when each $f_{k}$ is $C^1,$ $\sum_{k=1}^{\infty} f_{k}$ converges pointwise and $\sum_{k=1}^{\infty} f_{k}'$ converges uniformly.