I was making some limits using Maclaurin’s series and one of those was:
$$\lim _{x\to 0} \frac{e^x-\ln\left(1+x\right)-1}{x^2}$$
While this limit can be solved using Maclaurin’s series or L’Hospital’s rule, I want to find out a solution without using these methods—take for example, using just the rules of algebra of limits or the squeeze theorem. I am stuck with finding a solution. Any ideas?
If you are willing to take the convergence of the limit for granted, then we can determine the value of the limit: Define the function $f$ by
$$f(x) = \frac{e^x - 1 - \ln(1+x)}{x^2}$$
and suppose that $f(x)$ converges to some real number $L$ as $x \to 0$. Then $f$ satisfies the following identity:
\begin{align*} f(2x) = \frac{1}{2}f(x) + \frac{1}{4} \left( \frac{e^x - 1}{x} \right)^2 + \frac{1}{4(1+2x)}\frac{\ln\left(1+\frac{x^2}{1+2x}\right)}{\frac{x^2}{1+2x}}. \end{align*}
So by letting $x \to 0$ to both sides and using the known limits
$$ \lim_{x \to 0} \frac{e^x - 1}{x} = 1, \qquad \lim_{t \to 0} \frac{\ln(1+t)}{t} = 1, $$
we get
$$ L = \frac{1}{2} L + \frac{1}{4} + \frac{1}{4}. $$
Solving this equation for $L$, we conclude $L = 1$.
