Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it.
lim (tan(7x))^x
x~>0
To find the limit of (tan(7x))^x as x approaches 0, let's first simplify the expression.
Recall that the tangent function can be written as sin(x)/cos(x). So we have:
lim (sin(7x)/cos(7x))^x
x~>0
Now, let's rewrite the expression using exponential notation:
lim [(sin(7x)/cos(7x))^x]
x~>0
Now, let's rewrite the expression using logarithms:
lim [e^(ln(sin(7x)/cos(7x)))^x]
x~>0
Using logarithm properties, we can rewrite this further:
lim [e^(x * ln(sin(7x)/cos(7x)))]
x~>0
Now, we can use l'Hospital's Rule to evaluate the limit. Recall that l'Hospital's Rule states that if we have an indeterminate form of 0/0 or ∞/∞, we can take the derivative of the numerator and denominator repeatedly until we get a determinate form.
Let's find the derivative of the numerator and denominator separately.
The derivative of ln(sin(7x)/cos(7x)) with respect to x can be found using the chain rule. We have:
d/dx ln(sin(7x)/cos(7x))
= 1/(sin(7x)/cos(7x)) * (cos(7x) * 7 - sin(7x) * (-7))/cos^2(7x)
= 7/(sin(7x)/cos(7x)) * (cos(7x) + sin(7x))/cos^2(7x)
Now, let's substitute this derivative back into our limit expression:
lim [e^(x * (7*sin(7x) + 7*cos(7x))/(sin(7x) * cos(7x))^2)]
x~>0
Next, let's evaluate the limit as x approaches 0.
Since the numerator and denominator both approach 0 as x approaches 0, we have an indeterminate form of 0/0. Therefore, we can apply l'Hospital's Rule again.
Taking the derivative of the numerator and denominator again, we get:
(7*cos(7x) + 7*sin(7x))/(sin(7x) * cos(7x)) * (cos(7x) + sin(7x))/cos^2(7x)
Now, let's substitute this derivative back into our limit expression:
lim [e^(7*sin(7x) + 7*cos(7x))/(sin(7x) * cos(7x)) * (cos(7x) + sin(7x))/cos^2(7x)]
x~>0
At this point, we could continue to apply l'Hospital's Rule, but this approach seems to be quite complicated. So let's consider a more elementary method.
To simplify the expression, we can use the fact that sin(7x) and cos(7x) are both approximately equal to 0 when x is very close to 0.
lim (sin(7x) + cos(7x)) / (sin(7x) * cos(7x))
x~>0
Now, let's evaluate the limit as x approaches 0 by plugging in x = 0 into the expression:
(sin(7*0) + cos(7*0)) / (sin(7*0) * cos(7*0))
= (sin(0) + cos(0)) / (sin(0) * cos(0))
= (0 + 1) / (0 * 1)
= 1 / 0
= undefined
Therefore, the limit of (tan(7x))^x as x approaches 0 is undefined.
To find the limit of the given function, we can use l'Hospital's Rule if it is applicable. Let's first check if l'Hospital's Rule can be used in this case.
L'Hospital's Rule states that if we have a limit of the form "0/0" or "∞/∞", we can differentiate the numerator and denominator until we no longer have an indeterminate form.
In the given limit, we have the form "(tan(7x))^x". To see if it is of the form "0/0" or "∞/∞", we can evaluate the limit directly:
lim (tan(7x))^x
x~>0
When x approaches 0, the argument of the tangent function, 7x, also approaches 0. Therefore, we can rewrite the limit as:
lim [(sin(7x))/(cos(7x))]^x
x~>0
Now, let's evaluate the limit using l'Hospital's Rule. We differentiate the numerator and denominator separately and keep doing so until we no longer have an indeterminate form:
lim [(7*cos(7x))/(cos(7x)) * [(sin(7x))/(cos(7x))]^(x-1)
x~>0
Simplifying the expression inside the limit, we have:
lim 7 * [(sin(7x))/(cos(7x))]^(x-1)
x~>0
Now, we can try evaluating the limit directly without using l'Hospital's Rule:
lim 7 * (sin(7x)/cos(7x))^(x-1)
x~>0
As x approaches 0, the argument of the sine and cosine functions, 7x, also approaches 0. Using the limit properties, we can rewrite the expression as:
7 * (sin(0)/cos(0))^(0-1)
x~>0
Since sin(0) = 0 and cos(0) = 1, we have:
7 * (0/1)^(-1)
x~>0
Which simplifies to:
7 * (0)^(-1)
x~>0
Since any nonzero number raised to the power of -1 is equal to its reciprocal, we have:
7 * (∞)
x~>0
Which is equal to ∞.
Therefore, the limit of the given function, lim (tan(7x))^x as x approaches 0, is equal to ∞.
In this case, l'Hospital's Rule was not necessary to find the limit, as we could evaluate it directly by simplifying the expression.
You know that lim sinu/u = 1
so, lim tanu/u = 1 (because lim cosu = 1)
So, lim tan(7x)/x
= lim (tan(7x)*7)/(7x)
= 7 lim tanu/u where u7x
= 7
or, using L'Hospital,
lim tan(7x)/x
= lim (7 sec^2(7x))/1
= 7