I've been given the following assignment:
For $n \in\mathbb{N}\setminus\{0\}$, let $\phi_n:[1,\infty)\to \mathbb{R}$, given by $\phi_n(x):=2\dfrac{(2x)^{\frac{1}{n}}-1}{\frac{1}{n}-0}, \; x\in [1,\infty)$
One can show that there exists a $\phi:[1,\infty)\to\mathbb{R}$, such that $\phi_n$ converges to $\phi$ pointwise.
Show that $\phi_n|_{[1,4]}$ converges to $\phi|_{[1,4]}$ uniformly. Hint: Use the mean value theorem.
My work so far:
We can see that $\phi_n(x) = 2\cdot ((2x)^0)' = 2\cdot \ln(2x)$, so we have that $\phi:[1,\infty)\to\mathbb{R}$, is given by $x \mapsto 2\cdot\ln(2x)$ and now I want to show uniform convergence to $\phi|_{[1,4]}$, however, I haven't been able to come up with anything useful so far.
Thank you very much in advance.
Writing $f(t) = (2x)^{t}$, $\phi_n(x) = 2\frac{f(\frac{1}{n})-f(0)}{\frac{1}{n}}$,
then by MVT $\phi_n(x) = 2(2x)^{t} \ln(2x)$ for some $t \in (0, \frac{1}{n})$.
Now we want to bound $\phi(x)-\phi_n(x)$ uniformly in $x$, where $\phi(x) = 2 \ln(2x)$. We can say that $$|\phi_n(x) - \phi(x)| = \phi_n(x) - \phi(x)$$ $$ = 2 \ln(2x)(2^{t}-1) \leq 2 \ln(8)(2^{\frac{1}{n}}-1).$$ This is a uniform bound converging to zero in $n$.