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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.
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 1divided by the absolute value of x?
If you plot the graph of f(x) = 1/|x| in your graphing calculator and observe it, you'll see that the graph has a break in it (the line x = 0 (y-axis) is a vertical asymptote of the graph of the function), and is composed by two branches, which lie above the x-axis (because of the absolute value, and the line y = 0 is a horizontal asymptote).So that,as x approaches 0+ (from the right), f(x) increases without bound: lim(x→0+) 1/|x| = ∞, andas x increases without bound, the values of f(x) approaches 0: lim(x→+∞) 1/|x| = 0as x approaches 0- (from the left), f(x) increases without bound: lim(x→0+) 1/|x| = ∞, andas x decreases without bound, the values of f(x) approaches 0: lim(x→-∞) 1/|x| = 0Thus, lim(x→± ∞) 1/|x| = 0.By taking x close enough to 0, the values of f(x) do not approaches a number, so lim(x→0) 1/|x| does not exist (when we write symbolically lim(x→0) 1/|x| = ∞, we simply express the particular way in which the limit does not exist; 1/|x| can be made as large as possible as we like by taking x close enough to 0, but not equal to 0).
What is the limit of 1-cos x over x squared when x approaches 0?
1
What is the limit of sin3x times sin5x divided by x squared as x approaches 0?
For ease of writing, every time I say lim I mean the limit of _____ as x approaches zero. Lim sin3xsin5x/x^2=lim{[3sin(3x)/(3x)][5sin(5x)/(5x)]}. One property of limits is that lim sin(something)/(that same something)=1. So we now have lim{[3(1)][5(1)]}=15. lim=15
What is limit of 1 -cos x divided by x as x approaches 0?
1
How do you differentiate x over 2?
f(x) = x/2Then the differential is lim h->0 [f(x+h) - f(x)]/h= lim h->0 [(x+h)/2 - x/2]/h= lim h->0 [h/2]/h= lim h->0 [1/2] = 1/2f(x) = x/2Then the differential is lim h->0 [f(x+h) - f(x)]/h= lim h->0 [(x+h)/2 - x/2]/h= lim h->0 [h/2]/h= lim h->0 [1/2] = 1/2f(x) = x/2Then the differential is lim h->0 [f(x+h) - f(x)]/h= lim h->0 [(x+h)/2 - x/2]/h= lim h->0 [h/2]/h= lim h->0 [1/2] = 1/2f(x) = x/2Then the differential is lim h->0 [f(x+h) - f(x)]/h= lim h->0 [(x+h)/2 - x/2]/h= lim h->0 [h/2]/h= lim h->0 [1/2] = 1/2
How do you solve this Lim x - 0 x divided by x squared plus 2x?
Lim x (is going to 0) x/(x^2 + 2x)= Lim x (is going to 0) x/[x(x + 2)]= Lim x (is going to 0) 1/(x + 2) = 1/2
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 derivative of sin x?
f'[x] = lim(h->0) (f[x+h]-f[x])/h lim(h->0) (sin[x+h]-sin[x])/h By angle-addition formula, we have: lim(h->0) (sin[x]cos[h]+sin[h]cos[x]-sin[x])/h lim(h->0) (sin[x]cos[h]-sin[x])/h + lim(h->0) (sin[h]cos[x])/h sin[x]*lim(h->0) (cos[h]-1)/h + cos[x]*lim(h->0) sin[h]/h In a calculus class, it is shown that: lim(h->0) (cos[h]-1)/h = 0 and that lim(h->0) sin[h]/h is 1. So, sin[x]*lim(h->0) (cos[h]-1)/h + cos[x]*lim(h->0) sin[h]/h becomes sin[x]*0 + cos[x]*1 cos[x] So, if f[x] = sin[x], f'[x] = cos[x]
What is the Limit as X approaches infinity 5-X plus cosXoverX?
There are multiple ways to interpret this question-that is, you could mean either (5-x+cosx)/x or 5-x +(cosx/x). The limit of the second option is negative infinity because as x approaches infinity, |cosx/x|≤1, so 5-x+cosx/x is very close to 5-x, and 5-infinity is basically negative infinity. For the first option, we consider that -1≤cosx≤1. This implies that, as x approaches infinity, lim of (5-x-1)/x≤lim of (5-x+cosx)/x≤lim of (5-x+1)/x. Simplifying, we get that, as x approaches infinity, lim of (4-x)/x≤lim of (5-x+cosx)/x≤(6-x)/x. Simplifying our new limits, we get -1≤lim of (5-x+cosx)/x≤1. It is now clear that the limit of (5-x+cosx)/x as x approaches infinity =negative 1.
When the limit is approaching zero what is x divided by x squared plus 2x?
Lim x (is going to 0) x/(x^2 + 2x)= Lim x (is going to 0) x/[x(x + 2)]= Lim x (is going to 0) 1/(x + 2) = 1/2
What is the derivitive of 1?
The derivative of a constant is always 0. To show this, let's apply the definition of derivative. Recall that the definition of derivative is: f'(x) = lim h→0 (f(x + h) - f(x))/h Let f(x) = 1. Then: f'(x) = lim h→0 (1 - 1)/h = lim h→0 0/h = lim h→0 0 = 0!
