1. Statement of the problem: Prove the case of $\delta^+(f,x_0)=1$ for a real function $f$ at a point $x_0$ where $\delta^+$ denotes the right upper Dini derivative. 2. Formula and important rules. The definition is $$\delta^+(f,x_0)=\limsup_{h\to 0^+} \frac{f(x_0+h)-f(x_0)}{h}.$$ The meaning of $\limsup$ is that the limit of the supremum of the difference quotients as $h\to 0^+$ equals the given value. In plain language this means that for every $\varepsilon>0$ there are arbitrarily small positive $h$ with the quotient within $\varepsilon$ of $\delta^+$. 3. Proof when $\delta^+(f,x_0)=1$. Let $\varepsilon>0$ be arbitrary. By the definition of $\limsup$ there exists $\delta>0$ such that $\sup_{01-\varepsilon$. Hence there exists some $h$ with $01-\varepsilon$. Applying this with $\varepsilon=1/n$ for each positive integer $n$ we obtain a sequence $h_n\downarrow 0$ with $\frac{f(x_0+h_n)-f(x_0)}{h_n}>1-1/n$. Therefore the sequence of difference quotients satisfies $\lim_{n\to\infty} \frac{f(x_0+h_n)-f(x_0)}{h_n}=1$. This proves that when $\delta^+(f,x_0)=1$ there are arbitrarily small increments where the forward difference quotient approaches 1. 4. Conclusion. Thus the case $\delta^+=1$ yields the existence of a sequence $h_n\to 0^+$ with forward difference quotients tending to 1 and so the right upper Dini derivative equals 1 in the described sense.