What have I done wrong? Calculating $\lim\limits_{x\to 0}\frac{\ln(1+2x)}{x^2}$

Note that

$$\dfrac{\dfrac{2}{1+2x}}{2x}=\dfrac{2}{2x(1+2x)}.$$

Now

$$\lim_{x\to 0^-}\dfrac{2}{2x(1+2x)}=-\infty$$ and

$$\lim_{x\to 0^+}\dfrac{2}{2x(1+2x)}=+\infty.$$


Your second equality is not correct. Consider $\frac{1}{2}$. Divide $\frac{1}{2}$ by 2. It is not 1, but rather 1/4. Because

$\frac{1}{2}$ $\div$ 2 $=$ $\frac{1}{2}$ $(\frac{1}{2})$ $=$ $\frac{1}{4}$

Tags:

Limits

Analysis

Real Analysis

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