Question
10. \( \lim _{x \rightarrow 0} \frac{\sin 7 x}{x} \) 11. \( \lim _{x \rightarrow 0} \frac{\tan 3 x}{\sin 2 x} \)
Ask by Lane Mejia. in Morocco
Mar 09,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
Les limites sont 7 et 1.5.
Solution
Calculate the limit \( \lim_{x\rightarrow 0} \frac{\sin(7x)}{x} \).
Evaluate the limit by following steps:
- step0: Evaluate using L'Hopital's rule:
\(\lim _{x\rightarrow 0}\left(\frac{\sin\left(7x\right)}{x}\right)\)
- step1: Use the L'Hopital's rule:
\(\lim _{x\rightarrow 0}\left(\frac{\frac{d}{dx}\left(\sin\left(7x\right)\right)}{\frac{d}{dx}\left(x\right)}\right)\)
- step2: Find the derivative:
\(\lim _{x\rightarrow 0}\left(\frac{7\cos\left(7x\right)}{1}\right)\)
- step3: Simplify the expression:
\(\lim _{x\rightarrow 0}\left(7\cos\left(7x\right)\right)\)
- step4: Rewrite the expression:
\(7\times \lim _{x\rightarrow 0}\left(\cos\left(7x\right)\right)\)
- step5: Calculate:
\(7\times 1\)
- step6: Multiply:
\(7\)
Calculate the limit \( \lim_{x\rightarrow 0} \frac{\tan(3x)}{\sin(2x)} \).
Evaluate the limit by following steps:
- step0: Evaluate using L'Hopital's rule:
\(\lim _{x\rightarrow 0}\left(\frac{\tan\left(3x\right)}{\sin\left(2x\right)}\right)\)
- step1: Use the L'Hopital's rule:
\(\lim _{x\rightarrow 0}\left(\frac{\frac{d}{dx}\left(\tan\left(3x\right)\right)}{\frac{d}{dx}\left(\sin\left(2x\right)\right)}\right)\)
- step2: Find the derivative:
\(\lim _{x\rightarrow 0}\left(\frac{3\sec^{2}\left(3x\right)}{2\cos\left(2x\right)}\right)\)
- step3: Rewrite the expression:
\(\frac{\lim _{x\rightarrow 0}\left(3\sec^{2}\left(3x\right)\right)}{\lim _{x\rightarrow 0}\left(2\cos\left(2x\right)\right)}\)
- step4: Calculate:
\(\frac{3}{\lim _{x\rightarrow 0}\left(2\cos\left(2x\right)\right)}\)
- step5: Calculate:
\(\frac{3}{2}\)
La limite de \( \frac{\sin(7x)}{x} \) lorsque \( x \) tend vers 0 est égale à 7.
La limite de \( \frac{\tan(3x)}{\sin(2x)} \) lorsque \( x \) tend vers 0 est égale à 1.5.
Answered by UpStudy AI and reviewed by a Professional Tutor
Like
error msg
Explain 
Simplify this solution Bonus Knowledge
The limit \( \lim_{x \rightarrow 0} \frac{\sin 7 x}{x} \) can be evaluated using the fundamental limit \( \lim_{u \rightarrow 0} \frac{\sin u}{u} = 1 \). Here you can make the substitution \( u = 7x \), leading to \( \frac{\sin(7x)}{x} = 7 \frac{\sin(7x)}{7x} \), which gives you \( 7 \cdot 1 = 7 \) as the final answer. For the second limit \( \lim_{x \rightarrow 0} \frac{\tan 3 x}{\sin 2 x} \), you can again use the small-angle approximations. As \( x \) approaches 0, both \( \tan(3x) \) and \( \sin(2x) \) approximate their angles. Therefore, the limit becomes \( \frac{\tan(3x)}{\sin(2x)} ≈ \frac{3x}{2x} = \frac{3}{2} \), yielding a final answer of \( \frac{3}{2} \).
