I've managed to find the remainder term for the linearization of $\exp(x)$ about $x=0$ in Lagrange form: $$ R_1(x)=\exp(θ_Lx)\frac 12x^2 \text{ ,where } θ_L∈[0,1]. $$
My question is how would I find the value of $θ_L$ when $x$ is tending to $0$? I've tried to rearrange the above equation given that $R_1(x):=f(x)-P_k(x)$ to make $θ_L$ the subject and then find the limit of the equation when $x$ tends to $0$ but that didn't work. Any help would be appreciated.
Not sure if I understand the question, so I ask the OP to read the following and clarify the question accordingly. (Update: see the end of the post for the likely answer).
Lagrange remainder has the following (exact) form:
$$R_n=\int_{x_0}^x f^{(n+1)}(t)\frac{(x-t)^n}{n!} dt$$
In this case $f(t)=e^t$ and $x_0=0$, so:
$$R_1=\int_{0}^x e^t(x-t) dt=e^x-1-x$$
Which is of course, trivially true, and doesn't help much.
However, we can use the Mean-Value theorem which states that if a function is differentiable and continuous on $[a,b]$, there's at least one number $c$ such that:
$$f'(c)=\frac{f(b)-f(a)}{b-a}$$
From this theorem the following conclusion follows for some function $g(x)$ which is integrable and doesn't change sign on $[a,b]$:
$$\int_a^b f(t)g(t)dt=f(c) \int_a^b g(t)dt$$
Here number $c \in [a,b]$ can be different than in the previous example (depending on $g(x)$). Getting back to $R_n$, we have:
$$R_n=f^{(n+1)}(c) \int_{x_0}^x \frac{(x-t)^n}{n!} dt=f^{(n+1)}(c) \frac{(x-x_0)^{n+1}}{(n+1)!}$$
Thus, we have an equation for our case:
$$R_1=e^x-1-x=e^c \frac{x^2}{2}$$
$$e^c=2 \frac{e^x-1-x}{x^2}$$
Obviously, $c=c(x)$.
Now the OP seems to want $\lim_{x \to 0} c(x)$.
$$\lim_{x \to 0} e^c=2 \lim_{x \to 0} \frac{e^x-1-x}{x^2}=1$$
$$\lim_{x \to 0} c=0$$
Just as it should be because for $x \to 0$, because the quadratic term close to $0$ should be $x^2/2$.
If for example, we wanted to find $c$ for $x=1$, then we would have:
$$e^c=2(e-2)$$
$$c=\log 2+\log(e-2)$$
And the remainder (again, trivially true):
$$R_1=2(e-2)\frac{1}{2}=e-2 $$
Edit
However, if the OP searches $c$ in another form:
$$c(x)=\theta(x) x$$
Then of course there's a way to find the limit for $\theta$ from the above:
$$\theta(x) x= \log \left(2 \frac{e^x-1-x}{x^2} \right)$$
The above follows from:
$$\lim_{x \to 0} \frac{1}{x} \log \left(2 \frac{e^x-1-x}{x^2} \right) =\lim_{x \to 0} \frac{1}{x} \log \left(2 \frac{x^2/2+x^3/6}{x^2} \right)=\lim_{x \to 0} \frac{1}{x} \log \left(1+\frac{x}{3} \right)=\frac{1}{3}$$