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Question Number 88098 by ar247 last updated on 08/Apr/20

$$\underset{x\to 0^{+}}{\lim }(\frac{1}{x}-\frac{1}{\sin x})$$

Commented by ar247 last updated on 08/Apr/20

$$pleaseexplainttome$$

Commented by mathmax by abdo last updated on 08/Apr/20

$$l\left(x\right)=\frac{1}{x}-\frac{1}{sinx}=\frac{sinx-x}{xsinx}\sim \frac{x-\frac{x^3}{6}-x}{x^2}=-\frac{x}{6}\Rightarrow {lim}_{x\to 0}l\left(x\right)=0$$

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