Key to "I need to solve this limit: limx→0cos⁡x1sin⁡x"

High School
Geometry
Trigonometry
Trigonometric equations and identities

prsategazd

Answered question

2021-12-30

I need to solve this limit:

lim_{x \to 0} {\cos x}^{\frac{1}{\sin x}}

Answer & Explanation

raefx88y

Beginner2021-12-31Added 26 answers

1.Call this function f(x)
2. Evaluate the limit
$L = lim_{x \to 0} \log f (x) = lim_{x \to 0} \log ({(\cos x)}^{\frac{1}{\sin x}}) = lim_{x \to 0} \frac{(\log \cos x)}{\sin x}$ - this is a 0/0 indeterminate form but you can apply L'Hôpital to get
$lim_{x \to 0} \frac{(\frac{1}{\cos x}) \sin x}{\cos x} = lim_{x \to 0} \frac{\sin x}{\cos^{2} x} = 0$
The answer is $e^{L} = e^{0} = 1$

sonorous9n

Beginner2022-01-01Added 34 answers

Hint:
$lim_{x \to 0} {(\cos x)}^{\frac{1}{\sin x}}$
$= lim_{x \to 0} {(1 - \sin^{2} x)}^{\frac{12}{\sin x}}$
$\begin{matrix} = (lim_{x \to 0} (1 - \sin^{2} x)^{- \frac{1}{\sin^{2} x}})^{- lim_{x \to 0} \frac{\sin x}{2}} \end{matrix}$

Vasquez

Expert2022-01-09Added 669 answers

When
$(\cos x)^{\frac{1}{\sin x}}$
is meant
$\begin{matrix} lim_{x \to 0} \cos x^{\frac{1}{\sin x}} \\ = e^{lim_{x \to 0} \frac{\ln \cos x}{\sin x}} \\ = e^{lim_{x \to 0} \frac{x \cdot \ln \cos x}{x \cdot \sin x}} \\ = e^{lim_{x \to 0} \frac{\ln x \cos x}{x}} \\ = e^{lim_{x \to 0} \frac{(\cos x - 1) \cdot \ln (\cos x + 1 - 1)}{(\cos x - 1) x}} \\ v = e^{lim_{x \to 0} \frac{\cos x - 1}{x}} = e^{0} = 1 \end{matrix}$

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