Resources for becoming good at evaluation of limits.

Question

The concept of limits is quite clear to me. I used Calculus Early Transcendentals by James Stewart for learning the basic concepts of limits and even solved the problems given in it but on the internet and in some exams we have some difficult problems of evaluation of limits. I can cite an example problem to illustrate myself more properly $$ \lim_{x \to 0}~~ [tan ~~({\pi/4 + x})]^{1/x}$$.

These questions involve some knowledge which is beyond my current knowledge of limits. I searched the internet and found that textbooks are meant and focuses more epsilon-delta definition and it's proofs and questions like these are just for competitive exams and therefore I was unable to find these limits laws (derived limit laws for solving problems) in well known books like Spivak Calculus, Thomas Calculus, James Stewart Calculus etc.

There is a thread on MSE where @DaveL.Renfro shares his lecture notes and it's just great, I want something like that (more information than that). The reasoning he has used was also very agreeable, so I request you all to please refer me resources as per my need.

Thank you. Any suggestion might be helpful.

Answer 1

I think I might have a solution for that question in particular:

$$L = \lim_{x\rightarrow0}[\tan(\frac{\pi}{4} + x)]^\frac{1}{x}$$

Let $a=\frac{1}{x}$

As $x\rightarrow 0$, $a\rightarrow\infty$

$$L = \lim_{a\rightarrow\infty}[\tan(\frac{\pi}{4} + \frac{1}{a})]^a$$ $$L = [\lim_{a\rightarrow\infty}\tan(\frac{\pi}{4} + \frac{1}{a})]^{\lim_{a\rightarrow\infty}a}$$ $$L = [\tan(\frac{\pi}{4} + 0)]^{\lim_{a\rightarrow\infty}a}$$ $$L = [1]^{\lim_{a\rightarrow\infty}a}$$ $$L = 1$$

Answer 2

I don't have any resource recommendations at hand, but I can give you a tip for dealing with limits of the form $\lim_{x\to a}f(x)^{g(x)}$: rewrite $f(x)^{g(x)}$ as $e^{g(x)\ln(f(x))}$. Because of the continuity of the exponential function, this will boil down the problem of evaluating $\lim_{x\to a}f(x)^{g(x)}$ to that of evaluating

$$\lim_{x\to a}g(x)\ln(f(x))$$

In the case of evaluating $\lim_{x\to 0}\left(\tan\left(\frac{\pi}{4}+x\right)\right)^{\frac{1}{x}}$ (the limit is $e^2$, as will be seen), it would be enough to compute

$$\lim_{x\to 0}\frac{\ln\left(\tan\left(\frac{\pi}{4}+x\right)\right)}{x}$$

Note that numerator and denominator both approach $0$ as $x\to 0$. Since

$$\lim_{x\to 0}\frac{\frac{d}{dx}\ln\left(\tan\left(\frac{\pi}{4}+x\right)\right)}{\frac{d}{dx}x}=\lim_{x\to 0}\frac{\frac{\sec^2\left(\frac{\pi}{4}+x\right)}{\tan\left(\frac{\pi}{4}+x\right)}}{1}=\frac{\sec^2\left(\frac{\pi}{4}+0\right)}{\tan\left(\frac{\pi}{4}+0\right)}=\frac{1}{\cos^2\left(\frac{\pi}{4}\right)}=2$$

it follows from L'Hôpital's Rule that

$$\lim_{x\to 0}\frac{\ln\left(\tan\left(\frac{\pi}{4}+x\right)\right)}{x}=2$$

Thus,

$$\lim_{x\to 0}e^{\frac{1}{x}\ln\left(\tan\left(\frac{\pi}{4}+x\right)\right)}=e^2$$