# I need to solve this limit: \lim_{x \to 0} \cos x^{\frac{1}{\sin

I need to solve this limit:
$\underset{x\to 0}{lim}{\mathrm{cos}x}^{\frac{1}{\mathrm{sin}x}}$
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raefx88y

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

###### Not exactly what you’re looking for?
sonorous9n

Hint:
$\underset{x\to 0}{lim}{\left(\mathrm{cos}x\right)}^{\frac{1}{\mathrm{sin}x}}$
$=\underset{x\to 0}{lim}{\left(1-{\mathrm{sin}}^{2}x\right)}^{\frac{12}{\mathrm{sin}x}}$
$\begin{array}{}=\left(\underset{x\to 0}{lim}\left(1-{\mathrm{sin}}^{2}x{\right)}^{-\frac{1}{{\mathrm{sin}}^{2}x}}{\right)}^{-\underset{x\to 0}{lim}\frac{\mathrm{sin}x}{2}}\end{array}$

###### Not exactly what you’re looking for?
Vasquez

When
$\left(\mathrm{cos}x{\right)}^{\frac{1}{\mathrm{sin}x}}$
is meant
$\begin{array}{}\underset{x\to 0}{lim}\mathrm{cos}{x}^{\frac{1}{\mathrm{sin}x}}\\ ={e}^{\underset{x\to 0}{lim}\frac{\mathrm{ln}\mathrm{cos}x}{\mathrm{sin}x}}\\ ={e}^{\underset{x\to 0}{lim}\frac{x\cdot \mathrm{ln}\mathrm{cos}x}{x\cdot \mathrm{sin}x}}\\ ={e}^{\underset{x\to 0}{lim}\frac{\mathrm{ln}x\mathrm{cos}x}{x}}\\ ={e}^{\underset{x\to 0}{lim}\frac{\left(\mathrm{cos}x-1\right)\cdot \mathrm{ln}\left(\mathrm{cos}x+1-1\right)}{\left(\mathrm{cos}x-1\right)x}}\\ v={e}^{\underset{x\to 0}{lim}\frac{\mathrm{cos}x-1}{x}}={e}^{0}=1\end{array}$