How to solve this limit involving sine and log? I've tried L'Hopital's Rule but the differentiated numerator involves cos(1/x) which does not exist when x approaches 0.
$$ \lim_{x\to 0^+} \frac{x^2sin\frac{1}{x}}{\ln(1+2x)}$$
 A: $$0\le\left|\frac{x^2\sin\frac{1}{x}}{\ln(1+2x)}\right|\le\frac{x^2}{\ln(1+2x)}\sim\frac{2x}{\frac{2}{1+2x}}=x(1+2x)\to 0$$, where $\sim$ is the result of applyin L'Hospital.
A: Note that, by applying the Squeeze theorem you get
\begin{align}
&-1 \le \sin x \le 1\\
&\Rightarrow \lim_{x \to \infty} \frac{-1}{x} \le \lim_{x \to \infty} \frac{\sin x}{x} \le \lim_{x \to \infty} \frac{1}{x}\\
&\Rightarrow 0 \le \lim_{x \to \infty} \frac{\sin x}{x} \le 0\\
&\Rightarrow \lim_{x \to \infty} \frac{\sin x}{x} = 0\\
&\Rightarrow \lim_{y \to 0+} y \sin \frac{1}{y} = 0
\end{align}
so
\begin{align}
\lim_{x\to 0^+} \frac{x^2\sin\frac{1}{x}}{\ln(1+2x)} &= \lim_{x\to 0^+} \sin\left( \frac{1}{x} \right) \frac{x^2 \cdot 2x}{\ln(1+2x) \cdot 2x}\\
&= \lim_{x\to 0^+} \sin\left( \frac{1}{x} \right) \frac{x}{2}\\
&= 0\\
\end{align}
A: l'Hospital should also work. Just recall, that $\lim\text{(bounded)}\times\text{(converging to zero)} = 0$.
A: We know
$$\lim_{x\rightarrow 0} \ x\sin\left(\frac{1}{x}\right) = 0$$
$$\ln(1+2x) = 2x - \dfrac{4x^2}{2} + \dfrac{8x^3}{3} ... $$
So the limit reduces to :
$$\lim_{x\rightarrow 0} \ \frac{x\sin\left(\frac{1}{x}\right)x}{2x - \dfrac{4x^2}{2} + \dfrac{8x^3}{3} ..}$$
$$ = \lim_{x\rightarrow 0} \ \frac{x\sin\left(\frac{1}{x}\right)}{2 - \dfrac{4x}{2} + \dfrac{8x^2}{3} ..} = 0$$
A: $$\lim _{x\to \:0+}\left(\frac{x^2\sin \left(\frac{1}{x}\right)}{\ln \left(1+2x\right)}\right)$$
Squeeze Theorem:
Let f, g and h be functions such that for all $\color{green}{\:x\in \left[a,\:b\right]}$(except possibly at the limit point c).
$\color{green}{f\left(x\right)\le h\left(x\right)\le g\left(x\right)}$.
Also suppose that,$\color{green}{\lim _{x\to c}f\left(x\right)=\lim _{x\to c}g\left(x\right)=L}$.
Then for any$\color{green}{a\le c\le b,\:\lim _{x\to c}h\left(x\right)=L}$
Then Apply the Squeeze Theorem:
$-1\le \sin \left(\frac{1}{x}\right)\le \:1$
$$\lim _{x\to \:0+}\left(\frac{x^2\left(-1\right)}{\ln \left(1+2x\right)}\right)\le \lim \:_{x\to \:0+}\left(\frac{x^2\sin \left(\frac{1}{x}\right)}{\ln \left(1+2x\right)}\right)\le \lim \:_{x\to \:0+}\left(\frac{x^21}{\ln \left(1+2x\right)}\right)$$
By Hopital's Rule:
$$\lim _{x\to \:0+}\left(\frac{x^2\left(-1\right)}{\ln \left(1+2x\right)}\right)=-\lim _{x\to \:0+}\left(\frac{2x}{\frac{2}{2x+1}}\right)=\color{red}{0}$$
$$\lim _{x\to \:0+}\left(\frac{x^21}{\ln \left(1+2x\right)}\right)=\lim _{x\to \:0+}\left(\frac{2x}{\frac{2}{2x+1}}\right)=\color{red}{0}$$
So by the squeeze theorem $$\lim _{x\to \:0+}\left(\frac{x^2\sin \left(\frac{1}{x}\right)}{\ln \left(1+2x\right)}\right)=\color{red}{0}$$
