## anonymous one year ago Compute the limit: $\lim_{x\to0}\frac{\ln(\tan x)-\ln x}{e^{x^2}-1}$ Oh, and just for fun: no L'Hopital's rule allowed.

1. anonymous

Familiar, isn't it? @IrishBoy123

2. IrishBoy123

indeed! looks a lot scarier too, but merging the top and working with Maclaurin would be my way in i'll type my attempt in later.

3. IrishBoy123

$\lim_{x\to0}\frac{\ln(\tan x)-\ln x}{e^{x^2}-1}$ $=\lim_{x\to0}\frac{\ln \frac{\tan x}{ x}}{e^{x^2}-1}$ using Taylor expansions around x=0, for tan x, log x and e^x functions: $$\tan x = x+\frac{ x^3}{3}+\frac{2x^5}{15}-...$$ $$\frac{tan \ x}{x} = \frac{x+\frac{x^3}{3}+\frac{2x^5}{15}-...}{x} = 1 + \frac{x^2}{3} + O(x^4)$$ $$ln (\frac{\tan x}{ x}) = ln (1 + \frac{x^2}{3} + O(x^4))$$ $$ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} + ...$$ $$ln (1 + \frac{x^2}{3} ) = \frac{x^2}{3} - \frac{x^4}{18} + O(x^4)$$ $$e^x = 1 + x+ \frac{x^2}{2!} + \frac{x^3}{3!} + ...$$ $$e^{x^2} = 1 + x^2+ \frac{x^4}{2!} + \frac{x^6}{3!} + ...$$ $$e^{x^2} - 1 = x^2+ O(x^4)$$ $\lim_{x\to0}\frac{\ln \frac{\tan x}{ x}}{e^{x^2}-1}$ $= \lim_{x\to0} \frac{\frac{x^2}{3} - \frac{x^4}{18} + O(x^4)}{ x^2+ O(x^4)}$ $= \lim_{x\to0} \frac{\frac{1}{3} - \frac{x^2}{18} + O(x^2)}{ 1+ O(x^2)}$ $=\frac{1}{3}$

4. thomas5267

What are the conditions for the limit of the Taylor series equal to the limit itself?

5. Empty

Wait! Isn't the Maclauren series the same as differentiating essentially, so I am still curious about how you could do this without using L'Hopital's rule!

6. IrishBoy123

the proof of L'Hopital that i recall uses Maclaurin, so this all sounds very circular. begs the question, what is "the rule"?

7. IrishBoy123

maybe i've been carrying the wrong baggage around all this time but i figured that $\frac{f(x+\delta )}{g(x+\delta )}= \frac{f(x) + \delta f'(x) + ...}{g(x) + \delta g'(x) + ...}$ so f(x) = g(x) = 0 allows you to look at the derivatives meaning l'hopital's rule is just a black box? too hand wavey ?!

8. IrishBoy123

ouch, that looks like MAclaurin and makes your point but i can see it geometrically too |dw:1442128314117:dw|