Limit of a complicated function.

Question

Find $$\lim\limits_{x \to 2^{-}} \frac{e^{((x+2)\log 4){\frac{[x+1]}{4}}}-16}{ 4^x -16}$$ where $[x]$ denotes the greatest integer function less than or equal to x.

ATTEMPT:

I tried the following substitution $x=2-h$ and as $x \to 2$, $h \to 0.$ which gave me something like this

$\lim\limits_{h\to 0}$$ e^{((4-h)ln4){\frac{[3-h]}{4}}}-16\over \frac{16}{4^h} -16.$The denominator seems solvable as we get $\lim\limits_{h\to 0}$ $-16\frac{(4^h-1)h}{4^h*h}$ which seems a standard limit.But the numerator stills remain the same.

How can i simplify the numerator?

Answer 1

If we assume that there is a typo in the question (as mentioned in my comment) then (The mentioned typo has been fixed by OP)

We can write the given function as $$f(x) = \dfrac{\exp\left\{\{(x + 2)\log 4\}\cdot\dfrac{[x + 1]}{4}\right\} - 16}{4^{x} - 16}$$ When $x \to 2^{-}$ we can write $[x + 1] = 2$ and then the expression for $f(x)$ can be simplified to $$f(x) = \dfrac{\exp\left\{\left(1 + \dfrac{x}{2}\right)\log 4\right\} - \exp(2\log 4)}{4^{x} - 16}$$ Putting $x = 2 - h$ where $h \to 0^{+}$ we can see that the desired limit $L$ can be calculated as follows \begin{align} L &= \lim_{h \to 0^{+}}\dfrac{\exp\left\{\left(2 - \dfrac{h}{2}\right)\log 4\right\} - \exp(2\log 4)}{4^{2 - h} - 16}\notag\\ &= \exp(2\log 4)\lim_{h \to 0^{+}}\dfrac{\exp\left\{\left(2 - \dfrac{h}{2}\right)\log 4 - 2\log 4\right\} - 1}{\dfrac{16}{4^{h}} - 16}\notag\\ &= \frac{\exp(2\log 4)}{16}\lim_{h \to 0^{+}}4^{h}\cdot\dfrac{\exp\left\{\left(2 - \dfrac{h}{2}\right)\log 4 - 2\log 4\right\} - 1}{1 - 4^{h}}\notag\\ &= 1\cdot\lim_{h \to 0^{+}}1\cdot\dfrac{\exp\left\{\left(2 - \dfrac{h}{2}\right)\log 4 - 2\log 4\right\} - 1}{h}\cdot\frac{h}{1 - 4^{h}}\notag\\ &= -\frac{1}{\log 4}\lim_{h \to 0^{+}}1\cdot\dfrac{\exp\left\{\left(2 - \dfrac{h}{2}\right)\log 4 - 2\log 4\right\} - 1}{h}\notag\\ &= -\frac{1}{\log 4}\lim_{h \to 0^{+}}1\cdot\dfrac{\exp\left\{\left(2 - \dfrac{h}{2}\right)\log 4 - 2\log 4\right\} - 1}{\left(2 - \dfrac{h}{2}\right)\log 4 - 2\log 4}\cdot\dfrac{\left(2 - \dfrac{h}{2}\right)\log 4 - 2\log 4}{h}\notag\\ &= -\frac{1}{\log 4}\lim_{t \to 0^{-}}\frac{e^{t} - 1}{t}\cdot\left(-\frac{\log 4}{2}\right)\notag\\ &= \frac{1}{2} \end{align} In the above derivation we have used substitution $$t = \left(2 - \dfrac{h}{2}\right)\log 4 - 2\log 4$$ and $t \to 0^{-}$ when $h \to 0^{+}$. Also the following standard limits are used $$\lim_{x \to 0}\frac{e^{x} - 1}{x} = 1, \lim_{x \to 0}\frac{a^{x} - 1}{x} = \log a$$

Answer 2

First of all you can get rid of the denominator, by observing that your limit can be written $$ \lim_{x\to2^-}\frac{f(x)-16}{x-2}\frac{x-2}{4^x-16} $$ and the second factor is known to have limit $1/(4^2\log4)$, where $$ f(x)=\exp\left((x+2)\log 4\frac{[x+1]}{4}\right) $$ Now observe that, as $x\to2^-$, $[x+1]=2$, because you can restrict to be in a left neighborhood of $2$, so $2-\delta<x<2$ and $3-\delta<x+1<3$. By taking $\delta<1$, you get $[x+1]=2$.

Hence your limit is actually, apart from the factor you'll reinsert at the end $$ \lim_{x\to2^-}\frac{\exp\bigl((x+2)\log 4/2\bigr)-16}{x-2} $$ that's a simple derivative.