0
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
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What is the limit of x divided by e to the x as x approaches infinity?
Limit as x tends to ∞: x/e^xAs you can see, as x approaches infinity, the sum becomes ∞/∞. Now we use l'Hospitals rules.d/dx(x) = 1 (Derivative)d/dx(e^x) = e^x (Derivative)therefore, the sum can be written as lim x tends to ∞ 1/e^xNow as x approaches infinity, the sum = 1/∞ = 0Therefore, lim x tends to infinity: x/e^x = 0
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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.
How do you find vertical asymptote?
One way to find a vertical asymptote is to take the inverse of the given function and evaluate its limit as x tends to infinity.
What is the infinite limit of x - ln x?
Infinity
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x can go to + or - infinity. f(x) is limited from + 1/2 to - 1/2.
Is it generally true that limit of e to the power of fx as x tends to infinity is equal to e to the power of limit of fx as x tends to infinity?
In general, for a continuous function (one that doesn't make sudden jump - the type of functions you normally deal with), the limit of a function (as x tends to some value) is the same as the function of the limit (as x tends to the same value).e to the power x is continuous. However, you really can't know much about "limit of f(x) as x tends to infinity"; the situation may vary quite a lot, depending on the function. For example, such a limit might not exist in the general case. Two simple examples where this limit does not exist are x squared, and sine of x. If the limit exists, I would expect the two expressions, in the question, to be equal.
What is the answer to limit as x approaches infinity of e raised-x squared?
limit x tends to infinitive ((e^x)-1)/(x)
What is the limit of x divided by e to the x as x approaches infinity?
Limit as x tends to ∞: x/e^xAs you can see, as x approaches infinity, the sum becomes ∞/∞. Now we use l'Hospitals rules.d/dx(x) = 1 (Derivative)d/dx(e^x) = e^x (Derivative)therefore, the sum can be written as lim x tends to ∞ 1/e^xNow as x approaches infinity, the sum = 1/∞ = 0Therefore, lim x tends to infinity: x/e^x = 0
What is the limit as x approaches infinity of x plus x to the 3rd power plus x to the 5th power all divided by 1 - x to the 2nd power plus x to the 4th power?
Infinity.
What is an example of a calculus limit?
the limit [as x-->5] of the function f(x)=2x is 5 the limit [as x-->infinity] of the function f(x) = 2x is infinity the limit [as x-->infinity] of the function f(x) = 1/x is 0 the limit [as x-->infinity] of the function f(x) = -x is -infinity
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The answer depends on the side from which x approaches 0. If from the negative side, then the limit is negative infinity whereas if from the positive side, it is positive infinity.
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As x tends to negative infinity, the expression is asymptotically 0.
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1
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There is no maximum because y tends to + infinity as x tends to + or - infinity.
What is the limit as x approaches infinity of the square root of x?
What is the limit as x approaches infinity of the square root of x? Ans: As x approaches infinity, root x approaches infinity - because rootx increases as x does.
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.
Limit x approaches infinity for cos of x?
The limit does not exist.
