let $\phi (x) =\lim_{n \to \infty} \frac{x^n +2}{x^n +1}$; and $f(x) = \int_0^x \phi(t)dt$. Then $f$ is not differentiable at $1$. For $x \geq 0$, let $\phi (x) = \lim_{n \to \infty} \frac{x^n +2}{x^n +1}$; and $f(x) = \int_0^x \phi(t)dt$. Then $f$ is continuous at $1$ but not differentiable at $1$.
First we calculate $\phi (x) = \lim_{n \to \infty} \frac{x^n +2}{x^n +1}$.
For, $0 \leq x <1, \phi (x) = \lim_{n \to \infty} \frac{x^n +2}{x^n +1} = 2$ and for $ x \geq 1, \phi (x) = \lim_{n \to \infty} \frac{x^n +2}{x^n +1} = \phi (x) = \lim_{n \to \infty} \frac{1 +\frac{2}{x^n}}{1+\frac{1}{x^n}} =1 $.
Thus we have $f(x) = \int_0^x \phi(t)dt = \int_0^1 \phi(t)dt + \int_1^x \phi(t)dt = x+1$. Is the computation ok?
It can be easily verified that $f$ is continuous at $1$ but I have confusion about the other part that 'but not differentiable at $1$'.

First, we note that
$$\begin{align}
\phi(x) &= \lim_{n\to \infty} \frac{x^n+2}{x^n+1}\\\\
&=\begin{cases}
1,\,\,\text{for}\,\,x>1\\\\
\frac32 \,\,\text{for}\,\,x=1\\\\
2,\,\,\text{for}\,\,0\le x<1\\\\
\end{cases}
\end{align}$$

Next, let's form the difference quotients for $f$ at $x=1$.  Thus, for $h>0$, we have
$$\begin{align}
\frac{f(1+h)-f(1)}{h}&=\frac1h\int_1^{1+h}\phi(t)dt\\\\
&=\frac1h\left(\int_1^{1+h}1\,dt\right)\\\\
&=1
\end{align}$$
For $h <0$, 
$$\begin{align}
\frac{f(1+h)-f(1)}{h}&=\frac1h\int_1^{1+h}\phi(t)dt\\\\
&=\frac1h\left(\int_1^{1+h}2\,dt\right)\\\\
&=2
\end{align}$$
Inasmuch as the limit from the right
$$\lim_{h\to 0^{+}}\frac{f(1+h)-f(1)}{h}=1$$
while the limit from the left 
$$\lim_{h\to 0^{-}}\frac{f(1+h)-f(1)}{h}=2$$
are not equal, the limit does not exist.  Thus, $f'(1)$ does not exist as was to be shown!
