Question
31. \( \lim _{x \rightarrow 0} \frac{\sin x}{\operatorname{tg} x} \)
Ask by Mcfarlane Bartlett. in Democratic Republic of the Congo
Jan 13,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
La limite est 1.
Solution
Pour calculer la limite suivante :
\[
\lim_{x \rightarrow 0} \frac{\sin x}{\tan x}
\]
**Étapes de résolution :**
1. **Exprimer la tangente en termes de sinus et cosinus :**
\[
\tan x = \frac{\sin x}{\cos x}
\]
2. **Remplacer \(\tan x\) dans l'expression initiale :**
\[
\frac{\sin x}{\tan x} = \frac{\sin x}{\frac{\sin x}{\cos x}} = \cos x
\]
3. **Calculer la limite de \(\cos x\) lorsque \(x\) tend vers 0 :**
\[
\lim_{x \rightarrow 0} \cos x = \cos 0 = 1
\]
**Conclusion :**
La limite est donc égale à **1**.
Answered by UpStudy AI and reviewed by a Professional Tutor
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Simplify this solution The Deep Dive
First, let's rewrite the limit to simplify it. We know that \( \tan x = \frac{\sin x}{\cos x} \). Therefore, we can express the limit as follows: \[ \lim_{x \rightarrow 0} \frac{\sin x}{\tan x} = \lim_{x \rightarrow 0} \frac{\sin x}{\frac{\sin x}{\cos x}} = \lim_{x \rightarrow 0} \cos x \] As \( x \) approaches 0, \( \cos x \) approaches \( \cos(0) = 1 \). Thus, the limit evaluates to: \[ \lim_{x \rightarrow 0} \frac{\sin x}{\tan x} = 1 \] So, the final answer is: \[ \lim _{x \rightarrow 0} \frac{\sin x}{\tan x} = 1 \]
