(1)、求 $\lim _{x ightarrow 0} \frac{\sin x-\sin ( an x)}{x^{2} \ln (1+x)}$.

Question

(1)、求 $\lim _{x ightarrow 0} \frac{\sin x-\sin (	an x)}{x^{2} \ln (1+x)}$.

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ Evaluate the limit using L'Hôpital's Rule, which states that if the limit of functions $ f(x) $ and $ g(x) $ as $ x $ approaches a point $ c $ are both 0 or both $ \pm\infty $, then the limit of $ f(x)/g(x) $ as $ x $ approaches $ c $ is the same as the limit of their derivatives. ‖ ‖ step 2 ⋮ Differentiate the numerator $ \sin(x) - \sin(\tan(x)) $ and the denominator $ x^2 \ln(1 + x) $ with respect to $ x $. ‖ ‖ step 3 ⋮ Apply the derivatives and simplify to find the limit as $ x $ approaches 0. ‖ ‖ step 4 ⋮ The result of the limit is $ -1/3 $, as obtained from the Asksia-LL calculator. ‖ ∻Answer∻ ⚹ $ \lim _{x \rightarrow 0} \frac{\sin x-\sin (\tan x)}{x^{2} \ln (1+x)} = -\frac{1}{3} $ ⚹ ∻Key Concept∻ ⚹L'Hôpital's Rule and Taylor Series Expansion⚹ ∻Explanation∻ ⚹The limit is evaluated using L'Hôpital's Rule, which involves taking the derivative of the numerator and the denominator. Additionally, the Taylor series expansion is used to express the functions around the point $ x = 0 $ to find the limit.⚹◊What is the L'Hopital's rule and how can it be applied to find limits of trigonometric expressions like $\lim_{x \rightarrow 0} \frac{\sin x - \sin(\tan x)}{x^2 \ln(1+x)}$?⍭ Generate me a similar question◊

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