Solve Limit indeterminate Form using taylor Expansion

Question

It seems that the only way to remove the indeterminate form is to find equivalent of all the functions that pose the problem using taylor Expantion with the right order which is 3 in this case so we have

Taylor expansion for sqrt(1 + log(1+x) is

And for sinx ^2:

And so on to find a simplified function that goes to 0 can we use hopital rule here?

Answer 1

using $\ln(1+x) = x - x^2/2 + \cdots,$ you have $$\sqrt{1 + \ln(1+x)} = 1 + 1/2(x - x^2/2+\cdots) -1/8(x+\cdots)^2 + \cdots = 1 + \frac{1}{2}x - \frac{3}{8}x^2+\cdots \tag 1$$ and $$ 1 - \cos x - e^{x/2} = 1 - (1 - x^2/2+\cdots ) -(1+x/2 + x^2/8+\cdots) = -\frac{x}{2} + \frac{3}{8}x^2+\cdots \tag 2$$

$$ \sin^2 x = x^2 + \cdots \tag 3$$

adding $(1)$ and $(2)$ and using $(3)$, you get $$\lim_{x \to 0}\dfrac{\ln(1+x) + 1 - \cos x - e^{x/2}}{\sin^2x} = 0.$$

Answer 2

It's true that after L'Hopitals once, you get another indeterminate form. But if you use L'Hopitals a second time, you get $\dfrac 0 2$.

I recommend evaluating $\dfrac {\mathrm d^2}{\mathrm dx^2} \left({1 + \ln(1 + x)}\right)^{1/2}$ on its own so you don't make an algebra mistake, like I did the first one or two times. I'll let you confirm my algebra.