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The given limit is:

\[
\lim_{{x \to 0}} \left(1 + \sin x\right)^{\cot x}
\]

To evaluate this limit, we can use the exponential form to handle the indeterminate expression. Let's rewrite it as:

\[
\lim_{{x \to 0}} \left(1 + \sin x\right)^{\cot x} = e^{\lim_{{x \to 0}} \cot x \cdot \ln(1 + \sin x)}
\]

Now, we will evaluate the inner limit:

\[
\lim_{{x \to 0}} \cot x \cdot \ln(1 + \sin x)
\]

We know that \(\cot x = \frac{\cos x}{\sin x}\) and, as \(x \to 0\), \(\sin x \approx x\) and \(\cos x \approx 1\), so \(\cot x \approx \frac{1}{x}\). Also, for small values of \(x\), \(\ln(1 + \sin x) \approx \sin x = x\).

Thus, the limit simplifies to:

\[
\lim_{{x \to 0}} \frac{\ln(1 + \sin x)}{\sin x} \approx \lim_{{x \to 0}} \frac{x}{x} = 1
\]

Therefore, the exponential form of the limit is:

\[
e^{\lim_{{x \to 0}} \cot x \cdot \ln(1 + \sin x)} = e^{1} = e
\]

The final answer is:

\[
\lim_{{x \to 0}} \left(1 + \sin x\right)^{\cot x} = e
\]

Would you like more details or have any questions?

Here are 5 related questions for further exploration:
1. How would you evaluate the limit of \((1 + \sin x)^{1/x}\) as \(x \to 0\)?
2. What is the general approach to solving limits involving exponential forms?
3. How does L'Hospital's Rule apply to limits like this?
4. How do you determine when a limit tends towards \(e\) in indeterminate forms?
5. What are the conditions under which \(\lim_{{x \to 0}} \left(1 + x\right)^{\frac{1}{x}} = e\)?

**Tip:** For limits involving forms like \(0^0\), \(1^\infty\), or \(\infty^0\), rewriting the expression in exponential form is a helpful strategy.

Learn how to evaluate the limit lim(x→0) (1 + sin(x))^cot(x) using exponential forms and approximations. This problem covers trigonometric limits, logarithmic approximations, and exponential limit evaluation strategies. Suitable for undergraduate calculus students.

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