How the following limits can be computed without using Taylor series, Laurent series, or L'Hospital?
I) $$\lim_{x\to\pi/4} \frac{\ln\left(\tan\left(x\right)\right)}{\cos\left(2x\right)}$$
II) $$\lim_{x\to a} \frac{a^{x}-x^{a}}{x-a}$$
III)
$$\lim_{x\to 0} \frac{\ln\left(1+3^{x}\right)}{\ln\left(1+2x\right)}$$
IV) $$\lim_{x\to +\infty} \frac{x^{n}}{a^{x}}$$ my try:
about the first one :
since $\ln\left(x\right)$ is continuous over its domain hence the limit can be written as:
$$=\ln\left(\lim_{x\to\pi/4}\tan\left(x\right)^{^{\frac{1}{\cos\left(2x\right)}}}\right)=\ln\left(e^{\lim_{x\ to\pi/4}\large{\frac{\left(\tan\left(x\right)-1\right)}{\cos\left(2x\right)}}}\right)$$
this is where do I have a problem, how can I compute the limit just by some elementary ways? for the second and forth one I tried to use substitution but that does not work.
about the second case I applied the same strategy used for the first one,but I have the same problem
For the first limit, it can be made much simpler, with only high-school tools:
Set $t=\tan x\;$ ($t\to 1$ as $x\to\frac\pi 4$), so that the limit becomes, using some trigonometry, the limit, as $t\to 1$, of: $$\frac{\ln\left(\tan\left(x\right)\right)}{\cos\left(2x\right)}= \frac{\ln t}{\cfrac{1-t^2}{1+t^2}}=\frac{(1+t^2)\ln t}{1-t^2}=\underbrace{\frac{1+t^2}{1+t}}_{\downarrow\\1} \cdot\underbrace{\frac{\ln t}{1-t}}_{\;\downarrow\\-1}$$