L'hopital's question below:
lim x -> 0 from the positive direction of
( 5^(sin(x)) - 1 )/ (x)
To evaluate the limit using L'Hopital's Rule, we need to take the derivative of the numerator and the denominator separately.
First, let's start by finding the derivative of the numerator, which is (5^(sin(x)) - 1). The derivative of 5^(sin(x)) is found by applying the chain rule. The chain rule states that if we have a function f(g(x)), then the derivative is f'(g(x)) * g'(x). In this case, f(u) = 5^u and g(x) = sin(x). Therefore, the derivative of 5^(sin(x)) with respect to x is:
d/dx [ 5^(sin(x)) ] = ln(5) * 5^(sin(x)) * cos(x)
Next, let's find the derivative of the denominator, which is simply x:
d/dx [ x ] = 1
Now that we have the derivatives, we can rewrite the limit expression:
lim x -> 0+ ( 5^(sin(x)) - 1 )/ (x) = lim x -> 0+ ( ln(5) * 5^(sin(x)) * cos(x) )/ (1)
Since we still have an indeterminate form of 0/0, we can apply L'Hopital's Rule again by repeating the process of taking the derivatives of the numerator and denominator.
Taking the derivative of the numerator again:
d/dx [ ln(5) * 5^(sin(x)) * cos(x) ] = (-ln^2(5) * 5^(sin(x)) * sin^2(x)) + ln(5) * 5^(sin(x)) * cos^2(x)
And the derivative of the denominator:
d/dx [ 1 ] = 0
Now, we can rewrite the limit expression once again:
lim x -> 0+ ( ln^2(5) * 5^(sin(x)) * sin^2(x) - ln(5) * 5^(sin(x)) * cos^2(x) )/ (0)
Again, we have an indeterminate form of 0/0. We can repeat the process of taking the derivatives of the numerator and denominator until we no longer have an indeterminate form. However, in this case, it is not necessary to do so.
At this point, we have a limit expression in the form of 0/0, and taking the derivative again would yield the same result (0). Therefore, by applying L'Hopital's Rule twice, we've determined that the original limit evaluates to 0.