Evaluating limx→0(1sin⁡(x)−1x) without L'Hôpital's ruleThe only thing I can think of using is the basic identitylimx→0(sin⁡xx)=1but I can't reduce the original problem down to a point where I can apply this identity.

limx→0(1sin⁡x−1x)=limx→0(x−sin⁡xx2⋅xsin⁡x) =limx→0(x−sin⁡xx2) Now, use the Taylor series of sin⁡x

You could use the first few terms of the Taylor series for sin⁡(x) around x=0 (i.e., the Maclaurin series) in the numerator to get limx→0(1sin⁡x−1x)=limx→0(x−sin⁡xxsin⁡x)=limx→0(x−(x−x33+O(x5))xsin⁡x)=limx→0(x23+O(x4)sin⁡x)=limx→0(xsin⁡x)(x3+O(x3))=1(0)=0

limx→0(1sin⁡x−1x)=limx→0(x−sin⁡xx2⋅xsin⁡x) =limx→0(x−sin⁡xx2) Now, use the Taylor series of sin⁡x

You could use the first few terms of the Taylor series for sin⁡(x) around x=0 (i.e., the Maclaurin series) in the numerator to get limx→0(1sin⁡x−1x)=limx→0(x−sin⁡xxsin⁡x)=limx→0(x−(x−x33+O(x5))xsin⁡x)=limx→0(x23+O(x4)sin⁡x)=limx→0(xsin⁡x)(x3+O(x3))=1(0)=0

Key to "Evaluating limx→0(1sin⁡(x)−1x) without L'Hôpital's ruleThe only thing I..."

Jessie Lee

Answered question

2021-12-30

Evaluating

lim_{x \to 0} (\frac{1}{\sin (x)} - \frac{1}{x})

without L'Hôpital's rule
The only thing I can think of using is the basic identity

lim_{x \to 0} (\frac{\sin x}{x}) = 1

but I can't reduce the original problem down to a point where I can apply this identity.

Answer & Explanation

deginasiba

Beginner2021-12-31Added 31 answers

lim_{x \to 0} (\frac{1}{\sin x} - \frac{1}{x}) = lim_{x \to 0} (\frac{x - \sin x}{x^{2}} \cdot \frac{x}{\sin x})

= lim_{x \to 0} (\frac{x - \sin x}{x^{2}})

Now, use the Taylor series of

\sin x

Medicim6

Beginner2022-01-01Added 33 answers

Without LHopital

Vasquez

Expert2022-01-09Added 669 answers

You could use the first few terms of the Taylor series for $\sin (x)$ around x=0 (i.e., the Maclaurin series) in the numerator to get
$\begin{matrix} lim_{x \to 0} (\frac{1}{\sin x} - \frac{1}{x}) = lim_{x \to 0} (\frac{x - \sin x}{x \sin x}) \\ = lim_{x \to 0} (\frac{x - (x - \frac{x^{3}}{3} + O (x^{5}))}{x \sin x}) \\ = lim_{x \to 0} (\frac{\frac{x^{2}}{3} + O (x^{4})}{\sin x}) \\ = lim_{x \to 0} (\frac{x}{\sin x}) (\frac{x}{3} + O (x^{3})) \\ = 1 (0) \\ = 0 \end{matrix}$

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