To solve the given limit:

$\lim_{x \to 0} \left(\sin(3x)\right)^{1/x}$

Step-by-step solution:

1. Rewrite the limit: Take the natural logarithm to simplify the exponent: $y = \left(\sin(3x)\right)^{1/x} \implies \ln(y) = \frac{\ln(\sin(3x))}{x}.$

   Now, the limit becomes: $\ln(y) = \lim_{x \to 0} \frac{\ln(\sin(3x))}{x}.$

2. Simplify $\sin(3x)$: For small $x$, $\sin(3x) \approx 3x$. Substituting this approximation: $\ln(\sin(3x)) \approx \ln(3x) = \ln(3) + \ln(x).$

   Substituting into the limit: $\ln(y) = \lim_{x \to 0} \frac{\ln(3) + \ln(x)}{x}.$

3. Focus on $\ln(x)$: The term $\ln(x)$ dominates as $x \to 0$. Rewrite: $\ln(y) = \lim_{x \to 0} \frac{\ln(x)}{x}.$

   This limit diverges because $\ln(x) \to -\infty$ as $x \to 0^+$. Hence: $\ln(y) \to -\infty \implies y \to 0.$

4. Final result: Therefore: $\lim_{x \to 0} \left(\sin(3x)\right)^{1/x} = 0.$

Would you like a more detailed explanation or additional examples?

Related Questions:

1. What is the Taylor series expansion for $\sin(x)$ near $x = 0$?
2. How does the logarithmic approach help in solving exponential limits?
3. Can L’Hôpital’s Rule be applied directly to simplify this limit?
4. What happens if the base $\sin(3x)$ is replaced with $\cos(3x)$?
5. How do you evaluate similar limits involving logarithmic and trigonometric functions?

Tip:

For exponential functions involving limits, taking logarithms can often simplify the process dramatically by turning powers into products or quotients.