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Q: What is the infinite limit of x - ln x?
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What is the infinite limit of 1 divided by ln x?

The limit should be 0.


What is the infinite limit of ln 0?

If x --> 0+ (x tends to zero from the right), then its logarithm tends to minus infinity. On the other hand, x --> 0- (x tends to zero from the left) makes no sense, at least for real numbers, because the logarithm of negative numbers is undefined.


What is the derivative of y equals xlnx?

Use the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln xUse the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln xUse the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln xUse the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln x


Is there a function where the first derivative goes to infinity for x going to 0 and where the first derivative equals 0 when x is 1?

Yes, the function ln(x) where ln is the logarithm to base e.Yes, the function ln(x) where ln is the logarithm to base e.Yes, the function ln(x) where ln is the logarithm to base e.Yes, the function ln(x) where ln is the logarithm to base e.


What is Limit x to the power of x as x tends to infinity?

we can give a general expression: and limit is consider in only positive direction since ln eista for positives only nx is called the hyper power of x and when x tends to zero the general case is if n is a odd number then answer is zero if n is a even number it is 1 since consider the following example xx = ex ln(x) and when x tend s to zero the value is 1. let it is 3x = e x2 ln(x) whose value is zero similarly for other cases

Related questions

What is the infinite limit of 1 divided by ln x?

The limit should be 0.


What is the infinite limit of 1 divided by ln x divergent or convergent?

converges to zero (I think)


What is the infinite limit of ln 0?

If x --> 0+ (x tends to zero from the right), then its logarithm tends to minus infinity. On the other hand, x --> 0- (x tends to zero from the left) makes no sense, at least for real numbers, because the logarithm of negative numbers is undefined.


What is the derivative of y equals xlnx?

Use the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln xUse the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln xUse the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln xUse the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln x


How do you work out Ln 24 - ln x equals ln 6?

18


What is the dervative of x pwr x pwr x?

For the function: y = x^x^x (the superscript notation on this text editor does not work with double superscripts) To solve for the derivative y', implicit differentiation is needed. First, the equation must be manipulated so there are no x's raised to x's on the right side of the equation. So, both sides of the equation must be input into a natural logarithm, wherein we can use the properties of logarithms to remove the superscripted powers of the right side: ln(y) = ln(x^x^x) ln(y) = xxln(x) ln(y)/ln(x) = xx ln(ln(y)/ln(x)) = xln(x) eln(ln(y)/ln(x)) = exln(x) ln(y)/ln(x) = exln(x) ln(y) = ln(x)exln(x) Now there are no functions raised to functions (x's raised to x's). Deriving this equation yields: (1/y)(y') = ln(x)exln(x)(x(1/x) + ln(x)) + exln(x)(1/x) = ln(x)exln(x)(1 + ln(x)) + exln(x)(1/x) = exln(x)(ln(x)(1+ln(x)) + (1/x)) Solving for y' yields: y' = y[exln(x)(ln2(x) + ln(x) + (1/x))] or y = xx^x ln(y) = ln(x)x^x ln(y) = xxln(x) ln(y) = exlnxln(x) y'/y = exlnx[ln(x) + 1)ln(x) + exlnx(1/x) y' = y[exlnx(ln2(x) + ln(x) + 1/x)] y' = xx^x[exlnx(ln2(x) + ln(x) + 1/x)]


What is the derivative of x to the power of ln x?

I don't believe that the answer is ln(x)x^(ln(x)-2), since the power rule doesn't apply when you have the variable in the exponent. Do the following instead:y x^ln(x)Taking the natural log of both sides:ln(y)ln(x) * ln(x)ln(y) ln(x)^2Take the derivative of both sides, using the chain rule:1/y * y' 2 ln(x) / xy' 2 ln(x)/ x * yFinally, substitute in the first equation, y x^ln(x):y' 2 ln(x) / x * x^ln(x)y'2 ln(x) * x ^ (ln(x) - 1)Sorry if everything is formatted really badly, this is my first post on answers.com.


How do you solve for x 3 ln x - ln 3x equals 0?

3 ln(x) = ln(3x)ln(x3) = ln(3x)x3 = 3xx2 = 3x = sqrt(3)x = 1.732 (rounded)


Is there a function where the first derivative goes to infinity for x going to 0 and where the first derivative equals 0 when x is 1?

Yes, the function ln(x) where ln is the logarithm to base e.Yes, the function ln(x) where ln is the logarithm to base e.Yes, the function ln(x) where ln is the logarithm to base e.Yes, the function ln(x) where ln is the logarithm to base e.


Integrate of ln x squared?

int(ln(x2)dx)=xln|x2|-2x int(ln2(x)dx)=x[(ln|x|-2)ln|x|+2]


2 to the power of x equals 5 what is x?

2x = 5 then x*ln(2) = ln(5) so that x = ln(5)/ln(2) = 2.3219 approx.


How do you solve lntan(x)lnsin(x)-lncos(x)?

How do you solve ln|tan(x)|=ln|sin(x)|-ln|cos(x)|? Well you start by........