Question

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.

Answer 1

Familiar, isn't it? @IrishBoy123

Answer 2

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.

Answer 3

\[\lim_{x\to0}\frac{\ln(\tan x)-\ln x}{e^{x^2}-1}\]
\[=\lim_{x\to0}\frac{\ln \frac{\tan x}{ x}}{e^{x^2}-1}\]

using Taylor expansions around x=0, for tan x, log x and e^x functions:


\(\tan x = x+\frac{ x^3}{3}+\frac{2x^5}{15}-... \)

\(\frac{tan \ x}{x} = \frac{x+\frac{x^3}{3}+\frac{2x^5}{15}-...}{x} = 1 + \frac{x^2}{3} + O(x^4)\)

\( ln (\frac{\tan x}{ x}) = ln (1 + \frac{x^2}{3} + O(x^4)) \)

\(ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} + ...\)

\(ln (1 + \frac{x^2}{3} ) = \frac{x^2}{3} - \frac{x^4}{18} + O(x^4)\)

\(e^x = 1 + x+ \frac{x^2}{2!} + \frac{x^3}{3!} + ...\)

\(e^{x^2} = 1 + x^2+ \frac{x^4}{2!} + \frac{x^6}{3!} + ...\)

\( e^{x^2} - 1 = x^2+ O(x^4)\)

\[\lim_{x\to0}\frac{\ln \frac{\tan x}{ x}}{e^{x^2}-1}\]
\[= \lim_{x\to0} \frac{\frac{x^2}{3} - \frac{x^4}{18} + O(x^4)}{ x^2+ O(x^4)}\]
\[= \lim_{x\to0} \frac{\frac{1}{3} - \frac{x^2}{18} + O(x^2)}{ 1+ O(x^2)}\]
\[=\frac{1}{3}\]

Answer 4

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

Answer 5

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!

Answer 6

the proof of L'Hopital that i recall uses Maclaurin, so this all sounds very circular.

begs the question, what is "the rule"?

Answer 7

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 ?!

Answer 8

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