Whenever you have some long and difficult exponents, I recommend you to use natural logs to simplify the equation and then solve for the limit.
As y = (e^(7/x) + 4x)^(x/2), the natural logs of both sides of the equation would also be equal.
A = lny = ln[(e^(7/x) + 4x)^(x/2)].
***For the sake of simplicity, I will just set A as ln(y).
lim x -> 0+ [A] = lim x -> 0+ [ ln{(e^(7/x) + 4x)^(x/2)} ]
As you can move the exponent inside a natural log to the outside of the log, you can rearrange this equaion to this following:
lim x -> 0+ [A] = lim x -> 0+ [ (x/2) * ln{(e^(7/x) + 4x)} ]
The first method to use when computing limits is to directly substitute the value (0+). If we do so, the value would be 0 times natural log of infinity, meaning 0 times infinity, which is an indeterminate form, Remember, whenever we have an indeterminate form while computing a limit, we can use L'Hopital's rule. However, we got to transform the expression into a fraction form inorder to use the L'Hopital's rule.
You can simply transform this expression into a fraction like this:
lim x -> 0+ [A] = lim x -> 0+ [ [ln{(e^(7/x) + 4x)}] / (2/x) ]
***To make this clearer, the numerator of the fraction is ln{(e^(7/x) + 4x)} and the denominator is (2/x).
As this is an indeterminate form, we can solve the derivatives of both the numerator and the denominator to compute the limit.
The derivative of the numerator (ln{(e^(7/x) + 4x)}) is [(-7/(x^2))*e^(7/x) + 4] / (e^(7/x) + 4x).
The derivative of the denominator (2/x) is -2/x^2.
Now, we can put these back to the fraction and compute the limit.
lim x -> 0+ [A] = lim x -> 0+ [ ([(-7/(x^2))*e^(7/x) + 4] / (e^(7/x) + 4x)) / (-2/x^2)]
In order to simplify the fraction, we can multiply x^2 to both the numerator and the denominator. If we do that, the limit would look like this:
Afterwards, we can substitute 0+ to places of x that are appropriate (4x^2 and 4x) and be left with a limit like this:
*** I did not substitute 0+ to e^(7/x) as we want to cancel out these terms in later steps. If we do substitute over here, we would get another indeterminate form, which we would not like so!
lim x -> 0+ [A] = lim x -> 0+ [ -7*e^(7/x) / e^(7/x) / (-2) ]
And now, we can cancel out e^(7/x)
lim x -> 0+ [A] = lim x -> 0+ [ -7/ (-2) ] = lim x -> 0+ [7/2] = 7/2.
Last step! Remember that this value is the value of A and we were trying to solve for y. As lny = A, e^A = y. Therefore, the value of this limit (y) would be e^(7/2).
Apologies for the long response! Does this answer your question?
Hi James,
Thank you for your question.
Whenever you have some long and difficult exponents, I recommend you to use natural logs to simplify the equation and then solve for the limit.
As y = (e^(7/x) + 4x)^(x/2), the natural logs of both sides of the equation would also be equal.
A = lny = ln[(e^(7/x) + 4x)^(x/2)].
***For the sake of simplicity, I will just set A as ln(y).
lim x -> 0+ [A] = lim x -> 0+ [ ln{(e^(7/x) + 4x)^(x/2)} ]
As you can move the exponent inside a natural log to the outside of the log, you can rearrange this equaion to this following:
lim x -> 0+ [A] = lim x -> 0+ [ (x/2) * ln{(e^(7/x) + 4x)} ]
The first method to use when computing limits is to directly substitute the value (0+). If we do so, the value would be 0 times natural log of infinity, meaning 0 times infinity, which is an indeterminate form, Remember, whenever we have an indeterminate form while computing a limit, we can use L'Hopital's rule. However, we got to transform the expression into a fraction form inorder to use the L'Hopital's rule.
You can simply transform this expression into a fraction like this:
lim x -> 0+ [A] = lim x -> 0+ [ [ln{(e^(7/x) + 4x)}] / (2/x) ]
***To make this clearer, the numerator of the fraction is ln{(e^(7/x) + 4x)} and the denominator is (2/x).
As this is an indeterminate form, we can solve the derivatives of both the numerator and the denominator to compute the limit.
The derivative of the numerator (ln{(e^(7/x) + 4x)}) is [(-7/(x^2))*e^(7/x) + 4] / (e^(7/x) + 4x).
The derivative of the denominator (2/x) is -2/x^2.
Now, we can put these back to the fraction and compute the limit.
lim x -> 0+ [A] = lim x -> 0+ [ ([(-7/(x^2))*e^(7/x) + 4] / (e^(7/x) + 4x)) / (-2/x^2)]
In order to simplify the fraction, we can multiply x^2 to both the numerator and the denominator. If we do that, the limit would look like this:
lim x -> 0+ [A] = lim x -> 0+ [ [-7*e^(7/x) + 4x^2] / (e^(7/x) + 4x)) / (-2) ]
Afterwards, we can substitute 0+ to places of x that are appropriate (4x^2 and 4x) and be left with a limit like this:
*** I did not substitute 0+ to e^(7/x) as we want to cancel out these terms in later steps. If we do substitute over here, we would get another indeterminate form, which we would not like so!
lim x -> 0+ [A] = lim x -> 0+ [ -7*e^(7/x) / e^(7/x) / (-2) ]
And now, we can cancel out e^(7/x)
lim x -> 0+ [A] = lim x -> 0+ [ -7/ (-2) ] = lim x -> 0+ [7/2] = 7/2.
Last step! Remember that this value is the value of A and we were trying to solve for y. As lny = A, e^A = y. Therefore, the value of this limit (y) would be e^(7/2).
Apologies for the long response! Does this answer your question?