Find the limit without using L'Hospital's rule or show it doesn't exist

Question

This is the limit in question: $\lim_\limits{x\to 0}\frac{\ln \left(1+\sqrt{x\cdot \sin(x)}\right)}{x}$

I've tried making the substitution $\\ln \left (1+\sqrt{x\cdot \sin(x)}\right)=t$, but it didn't get me very far. I also know that the left and right hand limits aren't equal, but how to solve them and show this without L'Hospital?

Answer 1

\begin{aligned} \lim_{x\to 0^{+}}{\frac{\ln{\left(1+\sqrt{x\sin{x}}\right)}}{x}}&=\lim_{x\to 0^{+}}{\frac{\ln{\left(1+\sqrt{x\sin{x}}\right)}}{\sqrt{x^{2}}}}\\&=\lim_{x\to 0^{+}}{\sqrt{\frac{\sin{x}}{x}\times\frac{1}{x\sin{x}}}}\ln{\left(1+\sqrt{x\sin{x}}\right)}\\&=\lim_{x\to 0^{+}}{\sqrt{\frac{\sin{x}}{x}}\frac{\ln{\left(1+\sqrt{x\sin{x}}\right)}}{\sqrt{x\sin{x}}}}\\ &=\sqrt{1}\times 1\\ \lim_{x\to 0^{+}}{\frac{\ln{\left(1+\sqrt{x\sin{x}}\right)}}{x}}&=1 \end{aligned}

\begin{aligned} \lim_{x\to 0^{-}}{\frac{\ln{\left(1+\sqrt{x\sin{x}}\right)}}{x}}&=\lim_{x\to 0^{-}}{\frac{\ln{\left(1+\sqrt{x\sin{x}}\right)}}{-\sqrt{x^{2}}}}\\&=\lim_{x\to 0^{-}}{-\sqrt{\frac{\sin{x}}{x}\times\frac{1}{x\sin{x}}}}\ln{\left(1+\sqrt{x\sin{x}}\right)}\\&=\lim_{x\to 0^{-}}{-\sqrt{\frac{\sin{x}}{x}}\frac{\ln{\left(1+\sqrt{x\sin{x}}\right)}}{\sqrt{x\sin{x}}}}\\ &=-\sqrt{1}\times 1\\ \lim_{x\to 0^{-}}{\frac{\ln{\left(1+\sqrt{x\sin{x}}\right)}}{x}}&=-1 \end{aligned}

Because $ \lim\limits_{x\to 0}{\frac{\ln{\left(1+\sqrt{x\sin{x}}\right)}}{\sqrt{x\sin{x}}}}=\lim\limits_{y\to 0}{\frac{\ln{\left(1+y\right)}}{y}}=1 \cdot $

Answer 2

If you know the left and right limits aren't equal then you know that the limit at $x=0$ of the above function doesn't exist.

Is that what you're asking or are you looking for a proof that the left and right limits do not exist?