Function with two conditions I need a function that fulfills these two conditions:
$$
y'(0) = 1
$$
$$
\lim_{x\to\infty} y(x) = 70
$$
This results in a function that has two tangents: $y=x$ (blue) and $y=70$ (purple)
I have already figured out one function that fulfills both conditions: $y=\frac{70x}{x+70}$ (red)

The problem is that this function is too "slow" for my application. Is there any preferably simple equation that keeps the function's starting slope near 1 for a longer time so it "reaches" its limit (70) quicker? I made a quick drawing of what I mean:

The function will never receive negative $x$ values, so the part left of the y-axis doesn't matter. The function should never cross any of its tangents.
 A: Have you considered the arctangent function?  We can create a function in the form $$f(x)=a\arctan k(x-c)+d$$ which has a horizontal asymptote at $y=\dfrac{a\pi}{2}+d$.
Its derivative is $$f'(x)=\frac{ak}{1+[k(x-c)]^2}\implies f'(0)=\frac{ak}{1+k^2c^2}=1.$$
Solve for the appropriate parameters that you like.
A: How about 
$$  f(x) = \frac{70 \frac{1}{k} ((x+1)^k - 1)}{\frac{1}{k} ((x+1)^k - 1) + 70}  \text{?}  $$
Then 
$$  f'(x) = \frac{4900 k^2 (x+1)^{k-1}}{\left((x+1)^k+70 k-1\right)^2}  \text{,}  $$
so 
$$  f'(0) = 1  \text{.}  $$ 
For example, with $k = 2$:

$k$ need not be an integer.  For example, $k = \pi$.

A: Why don't you use an exponential?
$$f(x) = 70(1-e^{-\frac{x}{70}})$$
Then you would have
$$f'(x) = \frac{70}{70}e^{-x} = e^{-x} $$
And trivially $f'(0) = 1$. Bessides that
$$\lim_{x\to\infty} f(x) = 70\lim_{x\to\infty}(1-e^{-\frac{x}{70}}) =  $$
$$70(1-\lim_{x\to\infty}e^{-\frac{x}{70}}) = 70(1-0) = 70 $$
Above this all, $f$ has an exponential aproximation to the assymptote in $y = 70$
