Evaluate $\lim_{x \to 4} \frac{x^4-4^x}{x-4}$, where is my mistake? Once again, I am not interested in the answer. But rather, where is/are my mistake(s)? Perhaps the solution route is hopeless:
Question is: evaluate $\lim_{x \to 4} \frac{x^4 -4^x}{x-4}$.
My workings are:
Let $y=x-4$. Then when $x \to 4$, we have that $y \to 0$. Thus:
$$\lim_{y \to 0} \frac{(y+4)^4 - 4^{y+4}}{y} = \\ = \lim_{y \to 0}\frac{(y+4)^4}{y} - \lim_{y \to 0} \frac{4^{(y+4)}}{y}  $$
And this step is not allowed from the get go, as I am deducting infinities, which is indeterminate. What I should have done though:
$$4^4 \lim_{y \to 0} \frac{(1+y/4)^4-1+(4^y-1)}{y} = \\ 4^4 \lim_{y \to 0} \left( \frac{(1+y/4)^4-1}{\frac{y}{4}4} - \frac{4^y-1}{y} \right) = \\
=4^4\left(\frac{1}{4} \cdot 4 - \ln 4 \right) = 256(1-\ln 4)$$

 A: The step where you rewrite the limit as the difference of two other limits
((i) and (ii)) is not legitimate.
You can only equate a limit to a sum or difference of two limits
if both those limits converge.
A: The following step is not allowed
$$\lim_{x \to 4} \frac{x^4-4^x}{x-4}=\lim_{y \to 0} \frac{(y+4)^4-4^{y+4}}{y}\color{red}{=\lim_{y \to 0} \frac{(y+4)^4}{y}-\lim_{y \to 0} \frac{4^{y+4}}{y}}$$
Refer also to the related


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A: You can break up this limit under certain circumstances.
$\lim_\limits{x\to a} (f(x) + g(x)) = \lim_\limits{x\to a} f(x) + \lim_\limits{x\to a}g(x)$
You can do it if $\lim_\limits{x\to a} f(x), \lim_\limits{x\to a}g(x)$ both exist and are finite.
But if $\lim_\limits{x\to a} f(x) = \infty$ and $\lim_\limits{x\to a}g(x) = -\infty,$ then you have just given yourself and indeterminate form.
You could do this:
$\lim_\limits{y\to 0} \frac {(y+4)^4 - 4^{y+4}}{y} = \lim_\limits{y\to 0} \frac {y^4}{y}+ \lim_\limits{y\to 0} \frac {4y^3}{y}+\lim_\limits{y\to 0} \frac {6y^2}{y} +\lim_\limits{y\to 0} \frac {4y}{y} + \lim_\limits{y\to 0} \frac {4^4(1 - 4^y)}{y}$
Because each of those limits are defined and finite.
You could also use L'Hopital's rule.
