Answer :
To find the limit [tex]\(\lim _{x \rightarrow 0^{+}} x^{\sin x}\)[/tex], let's go through a detailed, step-by-step solution.
1. Understanding the Problem:
We are asked to find the limit as [tex]\( x \)[/tex] approaches [tex]\( 0^+ \)[/tex] for the expression [tex]\( x^{\sin x} \)[/tex]. Since [tex]\( \sin x \)[/tex] oscillates but tends to 0 as [tex]\( x \)[/tex] approaches 0, we need to handle the expression carefully to find the limit.
2. Rewrite the Expression:
Rewrite [tex]\( x^{\sin x} \)[/tex] using the exponential and natural logarithm functions:
[tex]\[ x^{\sin x} = e^{\ln(x^{\sin x})} = e^{\sin x \cdot \ln x} \][/tex]
This step uses the property that [tex]\( a^b = e^{b \ln a} \)[/tex].
3. Analyze the Exponent:
We now focus on the exponent [tex]\( \sin x \cdot \ln x \)[/tex]:
[tex]\[ \lim_{x \to 0^+} \sin x \cdot \ln x \][/tex]
As [tex]\( x \)[/tex] approaches [tex]\( 0^+ \)[/tex], [tex]\( \sin x \)[/tex] approaches 0. Meanwhile, [tex]\( \ln x \)[/tex] approaches [tex]\( -\infty \)[/tex].
4. Behavior of the Product:
Even though [tex]\( \ln x \)[/tex] diverges to [tex]\( -\infty \)[/tex] as [tex]\( x \)[/tex] approaches 0 from the positive side, the fact that [tex]\( \sin x \)[/tex] approaches 0 much faster suggests we might get an indeterminate form. To confirm, consider:
[tex]\[ \sin x \approx x \ \text{for} \ x \ \text{near 0}, \][/tex]
Thus, the product [tex]\( \sin x \cdot \ln x \)[/tex] can be approximated as [tex]\( x \cdot \ln x \)[/tex].
5. Limit of the Approximation:
We now evaluate:
[tex]\[ \lim_{x \to 0^+} x \ln x \][/tex]
Make a substitution to simplify:
Let [tex]\( t = \ln x \)[/tex]. As [tex]\( x \to 0^+ \)[/tex], [tex]\( t \to -\infty \)[/tex]. The substitution [tex]\( x = e^t \)[/tex] gives us:
[tex]\[ \lim_{t \to -\infty} t \cdot e^t \][/tex]
6. Evaluate the Simplified Limit:
[tex]\[ \lim_{t \to -\infty} t \cdot e^t = 0 \][/tex]
Since the exponential function [tex]\( e^t \)[/tex] decays to zero faster than [tex]\( t \)[/tex] grows negatively.
7. Return to Original Expression:
Therefore:
[tex]\[ \lim_{x \to 0^+} \sin x \ \ln x = 0 \][/tex]
8. Exponent Back to Exponential Form:
Using this result, we have:
[tex]\[ \lim_{x \to 0^+} x^{\sin x} = e^{\lim_{x \to 0^+} \sin x \cdot \ln x} = e^0 = 1 \][/tex]
Thus, the result of the limit is:
[tex]\[ \boxed{1} \][/tex]
1. Understanding the Problem:
We are asked to find the limit as [tex]\( x \)[/tex] approaches [tex]\( 0^+ \)[/tex] for the expression [tex]\( x^{\sin x} \)[/tex]. Since [tex]\( \sin x \)[/tex] oscillates but tends to 0 as [tex]\( x \)[/tex] approaches 0, we need to handle the expression carefully to find the limit.
2. Rewrite the Expression:
Rewrite [tex]\( x^{\sin x} \)[/tex] using the exponential and natural logarithm functions:
[tex]\[ x^{\sin x} = e^{\ln(x^{\sin x})} = e^{\sin x \cdot \ln x} \][/tex]
This step uses the property that [tex]\( a^b = e^{b \ln a} \)[/tex].
3. Analyze the Exponent:
We now focus on the exponent [tex]\( \sin x \cdot \ln x \)[/tex]:
[tex]\[ \lim_{x \to 0^+} \sin x \cdot \ln x \][/tex]
As [tex]\( x \)[/tex] approaches [tex]\( 0^+ \)[/tex], [tex]\( \sin x \)[/tex] approaches 0. Meanwhile, [tex]\( \ln x \)[/tex] approaches [tex]\( -\infty \)[/tex].
4. Behavior of the Product:
Even though [tex]\( \ln x \)[/tex] diverges to [tex]\( -\infty \)[/tex] as [tex]\( x \)[/tex] approaches 0 from the positive side, the fact that [tex]\( \sin x \)[/tex] approaches 0 much faster suggests we might get an indeterminate form. To confirm, consider:
[tex]\[ \sin x \approx x \ \text{for} \ x \ \text{near 0}, \][/tex]
Thus, the product [tex]\( \sin x \cdot \ln x \)[/tex] can be approximated as [tex]\( x \cdot \ln x \)[/tex].
5. Limit of the Approximation:
We now evaluate:
[tex]\[ \lim_{x \to 0^+} x \ln x \][/tex]
Make a substitution to simplify:
Let [tex]\( t = \ln x \)[/tex]. As [tex]\( x \to 0^+ \)[/tex], [tex]\( t \to -\infty \)[/tex]. The substitution [tex]\( x = e^t \)[/tex] gives us:
[tex]\[ \lim_{t \to -\infty} t \cdot e^t \][/tex]
6. Evaluate the Simplified Limit:
[tex]\[ \lim_{t \to -\infty} t \cdot e^t = 0 \][/tex]
Since the exponential function [tex]\( e^t \)[/tex] decays to zero faster than [tex]\( t \)[/tex] grows negatively.
7. Return to Original Expression:
Therefore:
[tex]\[ \lim_{x \to 0^+} \sin x \ \ln x = 0 \][/tex]
8. Exponent Back to Exponential Form:
Using this result, we have:
[tex]\[ \lim_{x \to 0^+} x^{\sin x} = e^{\lim_{x \to 0^+} \sin x \cdot \ln x} = e^0 = 1 \][/tex]
Thus, the result of the limit is:
[tex]\[ \boxed{1} \][/tex]
