Consider a function $f: \mathbb{R} \to \mathbb{R}$ that has an asymptote at $- \infty$ of the type $y=\lambda x + \beta$. According to trigonometry $\lambda=\tan{\theta}$ for a very small value of x at the point $P(k,f(k))$ near $- \infty$. So $$\lambda =\lim_{x \to -\infty} \frac{f(x)}{x} $$
Is it correct to state that $\lambda = \lim_{x \to -\infty} f'(x)$?
On
It is correct that $\lambda = \lim_{x\to-\infty}f'(x)$ in this situation but it's not true that $\lambda = f'(k)$ for some $k\in\mathbb R$.
Consider, for example, $f(x) = \frac{x^2}{x+1}$. This has an asymptote at $-\infty$ of slope $\lambda=1$. We have $f'(x) = 1 - \frac{1}{(x+1)^2}$, and so $f'(k)\neq 1$ for all $k\in\mathbb R$, since $\frac{1}{(k+1)^2}\neq 0$ for all $k$.
This example also shows that in general there is no $k$ with $\lambda = \frac{f(k)}{k}$: otherwise we would have $1 = \frac{x}{x+1}$, and multiplying this out we'd get $x = x+1$ which is impossible.
Yes, provided that the limit of the derivative exists.