I know, that applying standard linear approximation to $(1+x)^k$ at $0$ for any real k results in $1+kx$. However while reading Thomas' Calculus 14th Ed., in section 3.9 I encountered many examples using this approximation for values of $x$ near $0$ such as: $$\frac1{\sqrt{1-x^2}} \approx 1+{1\over2}x^2 \tag{1}$$ $$\sqrt[3]{1+5x^4} \approx 1+{1\over3}(5x^4) \tag{2}\label{2}$$
I saw many such examples on YouTube also. Let me work through the second example. I understand that for $x\approx0$, $5x^4$ is also approximately equal to zero and because of that one may fall into substituting $x$ in $1+kx$ with $5x^4$ and using it in approximation. But when I find the value of the derivative of \eqref{2} $$\frac{\mathrm{d}}{\mathrm{d}x}\sqrt[3]{1+5x^4}={1\over3}\left(1+5x^4\right)^{-2/3}(20x^3)$$ it's value at $x=0$ is $0$. Thus, applying std linear approximation to \eqref{2} near $0$ I get that it is approximately equal to $1+0x=1$. Hence, $1$ is the best linear approximation for \eqref{2}.
Why it is then many examples use the first approximation even when instead of $x$ there is some function of $x$ being used?
Because $1 + \tfrac{1}{3}(5x^4)$ is the the fourth order Taylor polynomial of $\sqrt[3]{1+5x^4}$ around $x=0$.
The Taylor series of a composition of functions such as $f(g(x))$ around $a$ can be found by plugging in the Taylor series of $g(x)$ around $a$ into the Taylor series of $f(x)$ around $g(a)$.
In your case we have $$f(x)=\sqrt[3]{1+x},$$ $$g(x) = 5x^4,$$ $$f(g(x)) = \sqrt[3]{1+5x^4}$$ and we are interested in the Taylor series of $f(g(x))$ around $0$.
Taking the Taylor series of both $f(x)$ and $g(x)$ around $0$ gives: $$f(x) = 1 + \tfrac{1}{3}x - \tfrac{1}{9}x^2 + \cdots,$$ $$g(x) = 5x^4,$$ where for $g(x)$ you can use the handy fact that The Taylor series of any polynomial is the polynomial itself. Plugging now the Taylor of $g(x)$ into $f(x)$ gives: $$f(g(x)) = 1 + \tfrac{1}{3}5x^4 - \tfrac{1}{9}(5x^4)^2 + \cdots,$$ Which we can truncate to $$f(g(x)) \approx 1 + \tfrac{1}{3}5x^4.$$