Evaluate $\lim_{x \to 4} \frac{x^4-4^x}{x-4}$, where is my mistake?

Question

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:

$$4^4 \lim_{y \to 0} \frac{(1+y/4)^4-1+(4^y-1)}{y} = \\ 4^4 \lim_{y \to 0} \left( \frac{(1+y/4)^4-1}{\frac{y}{4}4} - \frac{4^y-1}{y} \right) = \\ =4^4\left(\frac{1}{4} \cdot 4 - \ln 4 \right) = 256(1-\ln 4)$$

Answer 1

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

• Analyzing limits problem Calculus (tell me where I'm wrong).
• Evaluate $ \lim_{x \to 0} \left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right) $

Answer 2

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.

Answer 3

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.