limit of function $\sin(x \ln x)/x$ as $x\rightarrow 0$ I am trying to find $\lim \limits_{x \to 0} \frac{\sin(x \space \ln(x))}{x}$.
I believe I have solved it using the squeeze theorem to determine:
$\frac{-1}{x} \leq \frac{\sin(x \space ln(x))}{x} \leq ln (x)$, thus proving that the limit is -$\infty$.
However, our instructor has suggested that we approach this limit by multiplying by 1 (i.e. $\frac{f}{f}$) and using change-of-variable. Although my solution should suffice if correct, I was hoping someone could help me out with the lecturer's suggested approach. I am not sure what 1 to multiply the function by or what to choose as my change of variable. I have tried multiplying by $\frac{e^x}{e^x}$ and using both xln(x) and ln(x) in the change-of-variable with no luck.
 A: $$\lim \limits_{x \to 0} \frac{\sin(x \space \ln(x))}{x}=$$
$$\lim \limits_{x \to 0} \frac{\frac{d}{dx}\sin(\ln(x))}{\frac{dx}{dx}}=$$
$$\lim \limits_{x \to 0} \frac{(1+\ln(x))\cos(x\ln(x))}{1}=$$
$$\lim \limits_{x \to 0} (1+\ln(x))(\cos(x\ln(x))=$$
$$\lim \limits_{x \to 0} (1+\ln(x))\lim \limits_{x \to 0} \cos(x\ln(x))=$$
$$\lim \limits_{x \to 0} (1+\ln(x))\cos\left(\lim \limits_{x \to 0} x\ln(x)\right)=$$
$$\lim \limits_{x \to 0} (1+\ln(x))\cos\left(\lim \limits_{x \to 0}\frac{\ln(x)}{\frac{1}{x}}\right)=$$
$$\lim \limits_{x \to 0} (1+\ln(x)) \cos\left(\lim \limits_{x \to 0}\frac{\frac{d}{dx}\ln(x)}{\frac{d}{dx}\left(\frac{1}{x}\right)}\right)=$$
$$\lim \limits_{x \to 0} (1+\ln(x)) \cos\left(\lim \limits_{x \to 0}\frac{\frac{1}{x}}{-\frac{1}{x^2}}\right)=$$
$$\lim \limits_{x \to 0} (1+\ln(x)) \cos\left(\lim \limits_{x \to 0}-x\right)=$$
$$\lim \limits_{x \to 0}(1+\ln(x))\cos(0)=$$
$$\lim \limits_{x \to 0} 1+\lim \limits_{x \to 0}\ln(x)=$$
$$1+\lim \limits_{x \to 0}\ln(x)=$$
$$1+(-\infty)=-\infty$$
A: Rewrite as 
$$\lim\limits_{x\to 0} \frac{\sin(x\ln(x))\ln(x)}{x \ln(x)}$$
Note that $\lim\limits_{x\to 0}x\ln(x)=0$, this implies. 
$$\lim\limits_{x\to 0}\frac{\sin(x\ln(x))}{x\ln(x)}=\lim\limits_{u\to 0}\frac{\sin(u)}{u}=1$$
So then 
$$\lim\limits_{x\to 0} \frac{\sin(x\ln(x))\ln(x)}{x \ln(x)}=\lim\limits_{x\to 0}\ln(x)=-\infty$$
