Boilanubjaini8f

Answered

2022-06-25

Evaluate

$\underset{x\to 0}{lim}(\frac{1}{\mathrm{ln}(1+x)}+\frac{1}{\mathrm{ln}(1-x)})$

$\underset{x\to 0}{lim}(\frac{1}{\mathrm{ln}(1+x)}+\frac{1}{\mathrm{ln}(1-x)})$

Answer & Explanation

Josie123

Expert

2022-06-26Added 16 answers

What's wrong with L'Hopital's?

$\begin{array}{rl}\underset{x\to 0}{lim}\frac{1}{\mathrm{ln}(1+x)}+\frac{1}{\mathrm{ln}(1-x)}& =\underset{x\to 0}{lim}\frac{\mathrm{ln}(1-{x}^{2})}{\mathrm{ln}(1+x)\mathrm{ln}(1-x)}\\ & =\underset{x\to 0}{lim}\frac{-2x/(1-{x}^{2})}{\mathrm{ln}(1-x)/(1+x)-\mathrm{ln}(1+x)/(1-x)}\\ & =\underset{x\to 0}{lim}\frac{-2x}{(1-x)\mathrm{ln}(1-x)-(1+x)\mathrm{ln}(1+x)}\\ & =\underset{x\to 0}{lim}\frac{-2}{-1-\mathrm{ln}(1-x)-1-\mathrm{ln}(1+x)}\\ & =\frac{-2}{-2}\\ & =1.\end{array}$

$\begin{array}{rl}\underset{x\to 0}{lim}\frac{1}{\mathrm{ln}(1+x)}+\frac{1}{\mathrm{ln}(1-x)}& =\underset{x\to 0}{lim}\frac{\mathrm{ln}(1-{x}^{2})}{\mathrm{ln}(1+x)\mathrm{ln}(1-x)}\\ & =\underset{x\to 0}{lim}\frac{-2x/(1-{x}^{2})}{\mathrm{ln}(1-x)/(1+x)-\mathrm{ln}(1+x)/(1-x)}\\ & =\underset{x\to 0}{lim}\frac{-2x}{(1-x)\mathrm{ln}(1-x)-(1+x)\mathrm{ln}(1+x)}\\ & =\underset{x\to 0}{lim}\frac{-2}{-1-\mathrm{ln}(1-x)-1-\mathrm{ln}(1+x)}\\ & =\frac{-2}{-2}\\ & =1.\end{array}$

skylsn

Expert

2022-06-27Added 4 answers

Since $\mathrm{log}(1+x)=x-\frac{{x}^{2}}{2}+O({x}^{3})$, we have $\mathrm{log}(1-x)=-x-\frac{{x}^{2}}{2}+O({x}^{3})$. Therefore,

$\begin{array}{rl}\frac{1}{\mathrm{log}(1+x)}+\frac{1}{\mathrm{log}(1-x)}& =\frac{1}{x(1-\frac{x}{2}+O({x}^{2}))}+\frac{1}{-x(1+\frac{x}{2}+O({x}^{2}))}\\ \text{(}\ast \text{)}& & =\frac{1+\frac{x}{2}+O({x}^{2})}{x}-\frac{1-\frac{x}{2}+O({x}^{2})}{x}& =\frac{x+O({x}^{2})}{x}\end{array}$

where $(\ast )$ is because $\frac{1}{1+x}=1-x+O({x}^{2})$

Therefore,

$\underset{x\to 0}{lim}\frac{1}{\mathrm{log}(1+x)}+\frac{1}{\mathrm{log}(1-x)}=1$

$\begin{array}{rl}\frac{1}{\mathrm{log}(1+x)}+\frac{1}{\mathrm{log}(1-x)}& =\frac{1}{x(1-\frac{x}{2}+O({x}^{2}))}+\frac{1}{-x(1+\frac{x}{2}+O({x}^{2}))}\\ \text{(}\ast \text{)}& & =\frac{1+\frac{x}{2}+O({x}^{2})}{x}-\frac{1-\frac{x}{2}+O({x}^{2})}{x}& =\frac{x+O({x}^{2})}{x}\end{array}$

where $(\ast )$ is because $\frac{1}{1+x}=1-x+O({x}^{2})$

Therefore,

$\underset{x\to 0}{lim}\frac{1}{\mathrm{log}(1+x)}+\frac{1}{\mathrm{log}(1-x)}=1$

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