lim(x→0) [sin { x/n }+ cos { 3x/n } ]^( (2n)/x )
[tex]lim \: x→0 \: ( ({\sin( \frac{x}{n} )+ \cos( \frac{3x}{n} ) })^{ \frac{2n}{x} } ) [/tex]
Share
lim(x→0) [sin { x/n }+ cos { 3x/n } ]^( (2n)/x )
[tex]lim \: x→0 \: ( ({\sin( \frac{x}{n} )+ \cos( \frac{3x}{n} ) })^{ \frac{2n}{x} } ) [/tex]
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Answer:
To evaluate the limit as \(x\) approaches 0 for the expression \(\left(\sin\left(\frac{x}{n}\right) + \cos\left(\frac{3x}{n}\right)\right)^{\frac{2n}{x}}\), we can recognize this as an indeterminate form of \(1^\infty\). We can use L'Hôpital's Rule to simplify it.
Let \(f(x) = \sin\left(\frac{x}{n}\right) + \cos\left(\frac{3x}{n}\right)\) and \(g(x) = \frac{x}{2n}\) (inside the exponent).
Now, calculate the derivatives:
\[f'(x) = \frac{1}{n}\cos\left(\frac{x}{n}\right) - \frac{3}{n}\sin\left(\frac{3x}{n}\right)\]
\[g'(x) = \frac{1}{2n}\]
Now, we can rewrite the limit using L'Hôpital's Rule:
\[\lim_{{x \to 0}} \left(\sin\left(\frac{x}{n}\right) + \cos\left(\frac{3x}{n}\right)\right)^{\frac{2n}{x}} = \lim_{{x \to 0}} \frac{f(x)}{g(x)} = \frac{f'(0)}{g'(0)}\]
Substitute the derivatives:
\[\frac{\frac{1}{n}\cos(0) - \frac{3}{n}\sin(0)}{\frac{1}{2n}} = \frac{1}{\frac{1}{2}} = 2\]
So, \(\lim_{{x \to 0}} \left(\sin\left(\frac{x}{n}\right) + \cos\left(\frac{3x}{n}\right)\right)^{\frac{2n}{x}} = 2\).
Step-by-step explanation:
The given limit expression can be rewritten as:
[tex]\[ \lim_{{x \to 0}} \left( \sin\left(\frac{x}{n}\right) + \cos\left(\frac{3x}{n}\right) \right)^{\frac{2n}{x}} \][/tex]
Now, let's analyze the expression:
1. As [tex]( x \to 0 ),[/tex] both [tex]( \frac{x}{n} ) [/tex]and [tex]\( frac{3x}{n}) [/tex] approach [tex]( 0 ).[/tex]
2. The term [tex]( \sin(0) )[/tex] is \( 0 \) and [tex]( \cos(0)[/tex] is [tex]( 1 ).[/tex]
3. Therefore, the expression becomes [tex]( (0 + 1)^{\frac{2n}{x}}).[/tex]
Now, we have an indeterminate form of [tex]( 1^{\infty}).[/tex] To evaluate this, we can rewrite it as \[tex]( e^{\lim_{{x \to 0}} \frac{2n}{x} \cdot \ln(1)})[/tex], where [tex]( e )[/tex] is the base of the natural logarithm.
[tex]\[ \lim_{{x \to 0}} \frac{2n}{x} \cdot \ln(1) = \lim_{{x \to 0}} \frac{2n}{x} \cdot 0 = 0 \][/tex]
Therefore, the final limit is \( e^0 = 1 \).
So, [tex]\[ \lim_{{x \to 0}} \left( \sin\left(\frac{x}{n}\right) + \cos\left(\frac{3x}{n}\right) \right)^{\frac{2n}{x}} = 1 \][/tex]