Question
lim sin5-x/2x x->0 lim tan2x/sinx x->0 sachant que lim 1-4x/x² =1/2 calculer lim 1cos2x/3x² x->0
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Answer to a math question lim sin5-x/2x x->0 lim tan2x/sinx x->0 sachant que lim 1-4x/x² =1/2 calculer lim 1cos2x/3x² x->0
Sigrid
4.5119 Answers
1.
\lim_{{x \to 0}} \frac{\sin(5 - x)}{2x}
\sin(5 - x) \approx 5 - x \quad \text{(using small angle approximation)}
\lim_{{x \to 0}} \frac{5 - x}{2x} = \infty
So,\lim_{{x \to 0}} \frac{\sin(5 - x)}{2x} = \infty
2.
\lim_{{x \to 0}} \frac{\tan(2x)}{\sin(x)}
Using small angle approximations:
\tan(2x) \approx 2x \quad \text{and} \quad \sin(x) \approx x
\lim_{{x \to 0}} \frac{2x}{x} = 2
So,\lim_{{x \to 0}} \frac{\tan(2x)}{\sin(x)} = 2
3.
Using\lim_{{x \to 0}} \frac{1 - 4x}{x^2} = \frac{1}{2}
Find\lim_{{x \to 0}} \frac{1 - \cos(2x)}{3x^2}
Use the formula for small angle approximation:
\cos(2x) \approx 1 - 2x^2
1 - \cos(2x) \approx 2x^2
\lim_{{x \to 0}} \frac{2x^2}{3x^2} = \frac{2}{3}
So,\lim_{{x \to 0}} \frac{1 - \cos(2x)}{3x^2} = \frac{2}{3}
So,
2.
Using small angle approximations:
So,
3.
Using
Find
Use the formula for small angle approximation:
So,
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