$ \lim\limits_{x \to \infty} x^2(4^{\frac{1}{x}} - 4^{\frac{1}{1+x}}) $

Solution 1:

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

$$\frac{\log^2 4}{2} - \left(\frac{\log^2 4}{2} - \log 4\right) = \boxed{\log 4}$$

Solution 2:

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 $$\begin{align} \lim_{t\to 0}\frac{4^t-4^{t/(t+1)}}{t^2}&=\lim_{t\to 0}4^{t/(t+1)}\cdot\lim_{t\to 0}\frac{4^{t^2/(t+1)}-1}{\frac{t^2}{t+1}}\cdot\lim_{t\to 0}\frac{1}{t+1}\\ &=1\cdot \log 4\cdot 1=\log 4 \end{align}$$

Solution 3:

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 $$ \lim_{x\to\infty}x^2\big(4^{\frac{1}{x}}-4^{\frac{1}{x+1}}\big)= \lim_{x\to\infty}\frac{x^2}{x(x+1)}4^{\xi_x}\ln 4=\ln 4. $$