Using L'Hospital's Rule, find,
the limit as x approaches 0 of x^(sinx)
thank you in advance to any help.
To find the limit of the function as x approaches 0, you can use L'Hospital's Rule. This rule is applicable in cases where you have an indeterminate form, such as 0/0 or ∞/∞.
Let's start by differentiating the numerator and the denominator separately. We have the function f(x) = x^(sinx), and we want to find the limit of f(x) as x approaches 0.
Step 1: Take the natural logarithm of both sides of the expression to simplify the differentiation process.
ln(f(x)) = ln(x^(sinx))
Step 2: Apply the logarithmic property to bring the exponent down.
ln(f(x)) = (sinx) * ln(x)
Step 3: Differentiate both sides with respect to x using the product rule.
d/dx [ln(f(x))] = (cosx) * ln(x) + (sinx) * (1/x)
Step 4: Now, differentiate ln(f(x)) using the chain rule.
d/dx [ln(f(x))] = (1/f(x)) * f'(x)
Step 5: Substitute f'(x) with the right side from step 3.
(1/f(x)) * f'(x) = (1/f(x)) * [(cosx) * ln(x) + (sinx) * (1/x)]
Step 6: Simplify the expression.
1/f(x) = (cosx) * ln(x) + (sinx) * (1/x)
Step 7: Take the reciprocal on both sides to get f(x) = 1/{(cosx) * ln(x) + (sinx) * (1/x)}
Now, we have simplified the expression based on the rule of differentiation. At this point, we can evaluate the limit by substituting x = 0 into the function.
Taking the limit as x approaches 0:
lim(x->0) f(x) = lim(x->0) 1/{(cosx) * ln(x) + (sinx) * (1/x)}
This limit can be evaluated by applying L'Hospital's Rule again. Repeat the process of differentiating the numerator and denominator until you reach a solvable form.
Keep in mind that L'Hospital's Rule should be used iteratively if necessary, as long as the indeterminate form persists.