I am trying to solve the following L'hopital's limits question below:
lim x -> 0 from the positive direction of
( 5^(sin(x)) - 1 )/ (x)
then lim= n'/d' = cosx*ln5*5^sinx /1
lim= ln5 as x>0
graph the function and verify.
To solve this limit using L'Hopital's Rule, we need to find the derivative of the numerator and denominator separately.
First, let's find the derivative of the numerator. We have:
f(x) = 5^(sin(x)) - 1
To find the derivative, we can use the chain rule. The derivative of 5^(sin(x)) with respect to x is:
d/dx[5^(sin(x))] = ln(5) * 5^(sin(x)) * cos(x)
Now, let's find the derivative of the denominator:
g(x) = x
The derivative of x with respect to x is simply 1.
Now, we can take the limit of the derivatives.
lim x -> 0+ (from the positive direction)
lim x -> 0+ (ln(5) * 5^(sin(x)) * cos(x)) / 1
Now, we can evaluate this limit directly:
lim x -> 0+ (ln(5) * 5^(sin(0+)) * cos(0+)) / 1
Since cos(0+) = 1, we have:
lim x -> 0+ (ln(5) * 5^0 * 1) / 1
lim x -> 0+ (ln(5) * 1 * 1) / 1
lim x -> 0+ ln(5) = ln(5)
So, the limit as x approaches 0 from the positive direction of (5^(sin(x))-1)/x is ln(5).