Solve $\lim_{x\to 0} \frac{\ln(1+4x) - 4x}{x^2}$ without L'Hopital

Question

I tried solving $\lim_{x\to 0} \frac{\ln(1+4x) - 4x}{x^2}$ without L'Hopital like this:

$\lim_{x\to 0} \frac{\ln(1+4x)^{\frac {1}{4x}4x} - 4x}{x^2}$ = $\lim_{x\to 0} \frac{4x\ * \ln(e) - 4x}{x^2}$ = $\lim_{x\to 0} \frac{0}{x^2}$

and that is undetermined. And I have no other idea on how to solve it. Please help, and remember, WITHOUT L'Hopital rule. Solve just with simple limits manipulations, aka no Big O notations, series, integrals etc.

Can you use Taylor series? In particular $$\ln{(1+x)}\sim x-\frac12x^2\quad (\text{as }x\to0)$$ — Peter Foreman, Jan 27 '20 at 16:17
If you can use the sandwich theorem then try using $$x-\frac12x^2\le\ln{(1+x)}\le x-\frac12x^2+\frac13x^3$$ — Peter Foreman, Jan 27 '20 at 16:39
@PeterForeman, well your using of sandwich theorem is indeed based on the Taylor expansion so you've just changed the Taylor series into this form and it's not what can be used without any knowledge of power series. — , Jan 27 '20 at 17:44
https://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lhôpital-rule-or-series-expansion — lab bhattacharjee, Jan 27 '20 at 18:11
Thank, that answers my question completely, post it as an answer so I can give it a tick @labbhattacharjee — dzaralica69, Jan 28 '20 at 05:42

score 3 · Answer 1 · edited Jan 27 '20 at 18:25

3

replace $x$ by $-x$ & add then we get,

$2L=\lim_{x\to0}\dfrac{\ln(1-4x)+\ln(1+4x)}{x^2}=\lim_{x\to 0}\dfrac{\ln(1-16x^2)}{x^2}=-16\lim_{x\to 0}\dfrac{\ln(1-16x^2)}{-16x^2}\implies L=-8$

edited Jan 27 '20 at 18:25

Gary

31,845

answered Jan 27 '20 at 17:20

score 1 · Answer 2 · answered Jan 27 '20 at 16:22

1

Integrate $$ \frac{1}{{1 + t}} = 1 - t + \frac{{t^2 }}{{1 + t}} $$ from $0$ to $4x$ to obtain $$ \log (1 + 4x) = 4x - 8x^2 + \int_0^{4x} {\frac{{t^2 }}{{1 + t}}dt} \\ = 4x - 8x^2 + (4x)^3 \int_0^1 {\frac{{s^2 }}{{1 + 4xs}}ds} . $$ Hence $$ \frac{{\log (1 + 4x) - 4x}}{{x^2 }} = - 8 + 4^3 x\int_0^1 {\frac{{s^2 }}{{1 + 4xs}}ds} \to - 8 $$ as $x\to 0$.

answered Jan 27 '20 at 16:22

Gary

31,845

I can't use this because we haven't learnt it, I can use only limits manipulations or sandwich theorem. – dzaralica69 Jan 27 '20 at 16:33

Solve $\lim_{x\to 0} \frac{\ln(1+4x) - 4x}{x^2}$ without L'Hopital

I tried solving $\lim_{x\to 0} \frac{\ln(1+4x) - 4x}{x^2}$ without L'Hopital like this:

$\lim_{x\to 0} \frac{\ln(1+4x)^{\frac {1}{4x}4x} - 4x}{x^2}$ = $\lim_{x\to 0} \frac{4x\ * \ln(e) - 4x}{x^2}$ = $\lim_{x\to 0} \frac{0}{x^2}$

2 Answers2