I am trying to solve few questions though I know there answers I am not able to figure out the solution. I have tried to apply L'Hopital Rule as well as e^(ln) way for solving it but I am not obtaining desired answers. Your help will be much appreciated. Thank you. Here are the questions with their answers
1- $\lim_{x\to\inf}(1-\frac{4}{x-1})^{3x-1}$$=e^{-12}$
2-$\lim_{x\to2}\frac{2^x-x^2}{x^x-2^2}=\frac{log2-1}{log2+1}$
On
$(1-\frac{4}{x-1})^{3x-1}=(1-\frac{4}{x-1})^{3x-3}(1-\frac{4}{x-1})^{2}=((1-\frac{4}{x-1})^{x-1})^3(1-\frac{4}{x-1})^{2 }\to (e^{-4})^3 \cdot 1 = e^{-12}$ as $x \to \infty.$
What are your problems with L'Hopital's Rule ? This rule works for $ \lim_{x\to2}\frac{2^x-x^2}{x^x-2^2}$ !
On
$$\lim_{x\to2}\dfrac{2^x-x^2}{x^x-2^2}=\lim_{x\to2}\dfrac{2^x-2^2-(x^2-2^2)}{x^x-x^2+x^2-2^2}=\dfrac{2^2\cdot\lim_{x\to2}\dfrac{2^{x-2}-1}{x-2}-\lim_{x\to2}\dfrac{x^2-2^2}{x-2}}{x^2\cdot\lim_{x\to2}\dfrac{x^{x-2}-1}{x-2}+\lim_{x\to2}\dfrac{x^2-2^2}{x-2}}$$
$$=\dfrac{4\ln2-(2+2)}{4\ln2+(2+2)}$$
$$\lim_{x\to2}\dfrac{x^{x-2}-1}{x-2}=\dfrac{d(x^x)}{dx}_{\text{ at } x=2}$$
On
Use that
$$\left(1-\frac{4}{x-1}\right)^{3x-1}=\left[\left(1-\frac{4}{x-1}\right)^{\frac{x-1}{4}}\right]^{\frac{4(3x-1)}{z-1}}$$
and by $x=2+t, \quad t \to 0$
$$\frac{2^x-x^2}{x^x-2^2}=\frac{2^{t+2}-(t+2)^2}{(t+2)^{t+2}-2^2}=\frac{4\cdot 2^{t}-4+4-(t+2)^2}{(t+2)^{t+2}-4}=$$
$$=\left(4\cdot\frac{ 2^{t}-1}{t}+\frac{4-(t+2)^2}{t}\right)\frac{t}{(t+2)^{t+2}-4}\to\frac{4\log 2-4}{4\log 2+4}=\frac{\log 2-1}{\log 2+1}$$
indeed
$$\frac{(t+2)^{t+2}-4}{t}=\frac{e^{(t+2)\log(t+2)}-4}{t}=\frac{e^{2\log(t+2)}\cdot e^{t\log(t+2)}-e^{2\log(t+2)}+e^{2\log(t+2)}-4}{t}=$$
$$=(t+2)^2\frac{e^{t\log(t+2)}-1}{t\log(t+2)}\frac{t\log(t+2)}{t}+\frac{(t+2)^2-4}{t} \to4\log 2+4$$
The first one is a very standard limit. Note that $(1-\frac{4}{x-1})^{3x-1}=((1-\frac{1}{\frac{x-1}{4}})^{\frac{x-1}{4}})^{(3x-1)\frac{4}{x-1}}$. Can you finish from here?
For the second one just use L'Hopital rule once. To differentiate $x^x$ write it in the form $x^x=e^{xlnx}$.