Find the point where a normal to a function intersects the y-axis Question:
Let P be a point where the normal (in the point where the x-coordinate is h) to the curve
$$y = e^{2x} - 2x$$
cuts the y-axis. Determine the y-coordinates of P when h goes to 0.
Attempted solution:
I first decided to draw the following image:

So basically, we have the function for the curve, but we need to get on the normal and then move along the normal until it reaches the y-axis.
Let us start by taking the derivative of the function:
$$y' = 2e^{2x} - 2$$
The point x = h is the relevant point here, so:
$$y'(h) = 2e^{2h} - 2$$
Finding the k value of the normal:
$$k_1 k_2 = -1 \rightarrow k_2 = \frac{-1}{2e^{2h} - 2}$$
The equation for the normal is then:
$$y - (2e^{2h} - 2) = \frac{-1}{2e^{2h} - 2}(x-h)$$
Simplification gives:
$$y = 2e^{2h} - 2 -\frac{x-h}{2e^{2h} - 2}$$
If h goes to 0, this goes towards infinity. 
But that is hardly the case and cannot be true since it obviously has to cut the y-axis. Somewhere, something must have gone terribly, terribly wrong. The correct answer is $\frac{5}{4}$.
 A: The LHS in the equation of the normal should be $y - y(h)$, i.e. $y- e^{2h} - 2h$. Next, the point $P$ corresponds to $x =0$, so it has:
$$y_P = e^{2h} - 2h + \frac{h}{2e^{2h} - 2}$$
Using L'Hôpital's rule, we find: $y_P \to 1 + 1/4 = 5/4$
A: In the same spirit as MathInferno's answer, not using L'Hôpital's rule, consider Taylor series around $h=0$ $$e^{2h}= 1+2 h+2 h^2+O\left(h^3\right)$$ So, $$e^{2h}-2h=1+2 h^2+O\left(h^3\right)$$ $$2e^{2h}-2=4 h+4 h^2+O\left(h^3\right)$$ then $$e^{2h}-2h+\frac h{2e^{2h}-2}=1+2 h^2+O\left(h^3\right)+\frac h{4 h+4 h^2+O\left(h^3\right)}$$ $$e^{2h}-2h+\frac h{2e^{2h}-2}=1+2 h^2+O\left(h^3\right)+\frac 1{4 +4 h+O\left(h^2\right)}$$ Now, long division for the last term leads to $$e^{2h}-2h+\frac h{2e^{2h}-2}=\frac{5}{4}-\frac{h}{4}+O\left(h^2\right)$$ which shows the limit and how it is approached.
A: Here is how one would do the limit without L'Hôpital's rule for completeness:
$$\lim_{h \rightarrow 0} \left(e^{2h} -2h + \frac{h}{2e^{2h} - 2}\right) = 1 - 0 + \lim_{h \rightarrow 0} \frac{h}{2e^{2h} - 2} = 1 + \frac{1}{2} \lim_{h \rightarrow 0} \frac{h}{e^{2h} - 1}$$
$$\lim_{h \rightarrow 0} \frac{h}{e^{2h} - 1} = \frac{1}{2} \lim_{h \rightarrow 0} \frac{1}{\frac{e^{2h} - 1}{2h}}$$
Since $\frac{e^t -1}{t}$ is a standard limit equal to 1, the limit becomes $1 + \frac{1}{2} \cdot \frac{1}{2} = \frac{5}{4}$
A: @Claude Leibovici @Ahmed Hussein 
There is an alternative answer. 
Let us first recall that the envelope of the normals to a curve (called its evolute) is the locus of the centres of curvature for this curve.
Here, function $f$ is equivalent (see Remark 1) in the vicinity of $0$ to function $g$ defined by:
$$g(x)=1-\dfrac{x^2}{2}$$
(see graphics below where the curve of $f$ is black, and the curve of $g$ is red).
Let us switch then from the curve of $f$ to the equivalent curve $(C)$ of $g$, a parabola with summit $S=(0,1)$. (C) has clearly axis $Oy$ as its normal at point $S$.
The radius of curvature $R$ of curve (C) with equation $y=g(x)$ being the same as that of $y=f(x)$, the ordinate of the centre of curvature $P$ of (C) at point $S$ will be $R+1$. It suffices then to establish that $R=\dfrac{1}{4}$ for $x=0$.
This is immediate using classical formula
$$R=\dfrac{(1+g'(0)^2)^{3/2}}{|g''(0)|}$$
Remark 1 : Adjective "equivalent" may be ambiguous : the important thing is that $f^{(k)}=g^{(k)}$ for $k=0,1,2$.
Remark 2: the evolute of a parabola is represented here has a cusp.
Remark 3 : have a look at this remarkable blog of the AMS. Note that these curves are involutes ("building an/the involute" is the inverse operation of "building an/the evolute"). 

