$ \lim\limits_{x \to \infty} x^2(4^{\frac{1}{x}} - 4^{\frac{1}{1+x}}) $ I made a substitute $t = \frac{1}{x}$ now i have $ \lim\limits_{t \to 0} \frac{4^{t} - 4^{\frac{t}{1+t}}}{t^2} $ . I had the idea of adding $+1 -1$ like this $ \lim\limits_{t \to 0} \frac{(4^{t} - 1) -(4^{\frac{t}{1+t}})-1}{t^2} $ so I could seperate $\frac{4^{t}-1}{t}$ from the whole limit and try to reduce it to smaller one but I am still left with $\lim\limits_{t \to 0} \frac{1}{t}$ and the other half. Don't know if I could do anything more from this moment and I'm fresh out of ideas.
 A: Your add and subtract $1$ idea was a good one, however, that only works when the difference up top results in something of linear order like $t$. But when you do subtract the two terms in the numerator, you are left with something of quadratic order like $t^2$. Instead, add and subtract $1+t\log 4 $
$$\lim_{t\to0^+}\frac{(4^t-1-t\log 4 ) - (4^{\frac{t}{t+1}}-1-t\log 4 )}{t^2}$$
We can now split up the two limits. By the definition of the derivative, we have that
$$\lim_{x\to a}\frac{f(x)-f(a)-f'(a)(x-a)}{(x-a)^2} = \frac{1}{2}f''(a)$$
as long as the limit exists (which it will if the function is at least twice differentiable). This means the first limit evaluates to
$$\lim_{t\to0^+}\frac{4^t-1-t\log 4 }{t^2} = \frac{\log^2 4}{2}$$
For the second piece, use the substitution $s = \frac{t}{t+1} \implies t = \frac{s}{1-s}$
$$\lim_{s\to0^+}\frac{4^s-1-\frac{s\log 4}{1-s} }{\left(\frac{s}{1-s}\right)^2} = \lim_{s\to0^+}\frac{4^s-1-s\log 4 }{s^2}\cdot\frac{1}{(1-s)}-\frac{4^s-1}{s}\cdot\frac{1}{(1-s)^2}$$
$$= \frac{\log^2 4}{2}\cdot 1 - \log 4 \cdot 1 = \frac{\log^2 4}{2} - \log 4$$
which means the final answer is
$$\frac{\log^2 4}{2} - \left(\frac{\log^2 4}{2} - \log 4\right) = \boxed{\log 4}$$
A: Mean Value Theorem, for $f(x)=4^x$, with $f'(x)=4^x\ln 4$: There exists a $\xi_x\in\big(1/x,1/(x+1)\big)$, such that
$$
4^{\frac{1}{x}}-4^{\frac{1}{x+1}}=\left(\frac{1}{x}-\frac{1}{x+1}\right)4^{\xi_x}\ln 4=\frac{4^{\xi_x}\ln 4}{x(x+1)}
$$
Clearly $\lim_{x\to\infty}\xi_x=0$, and hence $\lim_{x\to\infty}4^{\xi_x}=1$. Hence
$$
\lim_{x\to\infty}x^2\big(4^{\frac{1}{x}}-4^{\frac{1}{x+1}}\big)=
\lim_{x\to\infty}\frac{x^2}{x(x+1)}4^{\xi_x}\ln 4=\ln 4.
$$
A: Write
$$\frac{4^t-4^{t/(t+1)}}{t^2}=4^{t/(t+1)}\cdot\frac{4^{t^2/(t+1)}-1}{\frac{t^2}{t+1}}\cdot\frac{1}{t+1} $$
and use the standard limit $\lim_{u\to 0}\frac{a^u-1}{u}=\log a$ to find
$$\begin{align}
\lim_{t\to 0}\frac{4^t-4^{t/(t+1)}}{t^2}&=\lim_{t\to 0}4^{t/(t+1)}\cdot\lim_{t\to 0}\frac{4^{t^2/(t+1)}-1}{\frac{t^2}{t+1}}\cdot\lim_{t\to 0}\frac{1}{t+1}\\
&=1\cdot \log 4\cdot 1=\log 4
\end{align}$$
A: Too long for a comment.
The limit itself has been given in the answers. However, we could go beyond the limit since we could look at the problem thinking about the loss of accuracy when working with limited precision.
To make the problem more general, consider
$$y=x^2(a^{\frac{1}{x}} - a^{\frac{1}{x+b}})-b \log(a)\qquad \text{with}\qquad b >0$$ Using the same steps as in other answers, we can show that
$$y=b \log (a)\Bigg[\frac{\log (a)-b}{x}+\frac{\log ^2(a)-3 b \log (a)+2 b^2}{2
   x^2}\Bigg]+O\left(\frac{1}{x^3}\right)$$
For a simple illustration, using $a=e^2$, $b=1$ and $x=10^{15}$, the above gives
$$y=2\times 10^{-15}$$ while computed with limited precision the result would be $0$.
