The evaluation of a limit or a definite integral almost every time requires a huge amount of algebra. Whenever we need to evaluate a limit of a nasty function we start doing algebra just hoping it to get converted into one of those standard limits. Converting a given limit into one of those known limits is vey tedious task and as Prof. David Jerison describes non-calculus step . Same thing is with the definite integrals, at elementary level we seek for a function whose antiderivative we already know.
For example $$ \lim_{x \to 0}~~ [tan(\pi/4 +x)]^{1/x}$$ $$ \lim_{x \to 0}\left [ \frac{1+tanx}{1-tanx} \right ]^{1/x}$$ Now I'm stuck. Although, it looks to me a little alike $$ \lim_{x\to0} (1+x)^{1/x}$$ whose value is $e$ but I don't know how to proceed after that (It's just a sample problem to illustrate myself as to what problem I'm getting). There many many instances in integral calculus where we have to do so much algebra just for converting it into one of those known forms.So, my question is
Why everyone has done it from past 300 years and got no problem but I'm getting no way out of it? How they did it? How you do it?
There is something called Taylor Expansion which converts almost every function into an infinite series and hence we can get a good approximation by evaluating the limits or integrals of Taylor Expansion of the given function.The expansion of $e^x$ is very useful in calculating limits and integrals involving it.
Is it clever to always go for Taylor Expansion? Can it always save us from unpredictable algebra?
I have gone for books which are usually recommended like Stewart Calculus, Thomas Calculus, Spivak Calculus etc. but these books don't say much about hard limits (they say very little on limits) and integrals (although Spivak Calculus is a little more rigorous on integrals).
This question of mine may seem to be like a rant (up to certain extent it is) and personal problems are usually discouraged on sites like these but I tried to make it as all-pervading as possible. The question is not How can I solve limits and integrals? BUT How is/was it solved by everyone else (living and gone by) knowing that it only involves converting one thing into something other?
Thank you. Any help will be much appreciated.
Just an illustration for the given problems.
Taylor series is, for sure, very good for finding not only limits but also how they are approached. The problem is that we very often need to compose Taylor series.
Considering $$y=\sqrt[x]{\tan \left(x+\frac{\pi }{4}\right)}\implies \log(y)=\frac 1x \log\left(\tan \left(x+\frac{\pi }{4}\right)\right)$$
Now start with $$\tan \left(x+\frac{\pi }{4}\right)=1+2 x+2 x^2+\frac{8 }{3}x^3+O\left(x^4\right)$$ Continue with Taylor to get $$\log\left(\tan \left(x+\frac{\pi }{4}\right)\right)=2 x+\frac{4 }{3}x^3+O\left(x^4\right)$$ which makes $$\log(y)=2 +\frac{4 }{3}x^2+O\left(x^3\right)$$ Continue again $$y=e^{\log(y)}=e^2+\frac{4 e^2 }{3}x^2+O\left(x^3\right)$$
If you prefer to write
$$y=\left(\frac{1+\tan (x)}{1-\tan (x)}\right)^{\frac{1}{x}}\implies \log(y)=\frac 1 x \log\left(\frac{1+\tan (x)}{1-\tan (x)}\right)$$ start with $$\tan(x)=x+\frac{1}{3}x^3+\frac{2 }{15}x^5+O\left(x^7\right)$$ which makes $$\frac{1+\tan (x)}{1-\tan (x)}=\frac{1+x+\frac{1}{3}x^3+\frac{2 }{15}x^5+O\left(x^7\right)}{1-x-\frac{1}{3}x^3-\frac{2 }{15}x^5+O\left(x^7\right)}$$ perform the long division to get $$\frac{1+\tan (x)}{1-\tan (x)}=2 x+\frac{4 }{3}x^3+O\left(x^4\right)$$ and just continue.