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Lim x approaches 0 x x x x-?
When the limit of x approaches 0 the degree on n is greater than 0.
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 2 of x² - 2x over x² - x - 2?
Lim(x→2) (x2 - 2x) / (x2 - x - 2) = Lim(x→2) x(x - 2) / (x - 2)(x + 1) = Lim(x→2) x / (x + 1) = 2/3
Limit when x goes to 0 tanx divided by x?
tan x = sin x / cos x, so:lim (tan x / x) = lim (sin x / x cos x). Since it is known that the limit of sin x / x = 1, you have lim 1 / cos x = 1 (since cos 0 = 1).tan x = sin x / cos x, so:lim (tan x / x) = lim (sin x / x cos x). Since it is known that the limit of sin x / x = 1, you have lim 1 / cos x = 1 (since cos 0 = 1).tan x = sin x / cos x, so:lim (tan x / x) = lim (sin x / x cos x). Since it is known that the limit of sin x / x = 1, you have lim 1 / cos x = 1 (since cos 0 = 1).tan x = sin x / cos x, so:lim (tan x / x) = lim (sin x / x cos x). Since it is known that the limit of sin x / x = 1, you have lim 1 / cos x = 1 (since cos 0 = 1).
Why the number 0 has no reciprocal?
Division by zero is not allowed/defined. So you cannot take 'one over zero', or have zero in the denominator.Without going too technical, a person might say that 1/0 is infinity, and it sounds good. But if you have a function [say f(x) = 1/x] and take the limit of f(x) as x approaches zero, then f(x) approaches infinity as x approaches from the right, but it approaches negative infinity as you approach from the left, therefore the limit does not exist.
