0
First, you can take the constant factor 3 out, to obtain 3 times the limit of (1 - cos x) / x.
Since this is of the form 0/0, you can use L'Hôpital's rule, which states that in such cases, you can take the derivative of both numerator and denominator. This results in the limit (as x approaches 0) of sin x / x, that is, 1 / 1 = 1. So, the final result is 3 times the limit of 1 = 3.
What else can I help you with?
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
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!
What is limit as x approaches 0 of sin squared x by x?
lim(x->0) of sin(x)^2/x we use L'Hospital's Rule and derive the top and the bottomd/dx(sin(x)^2/x)=2*sin(x)*cos(x)/1lim(x->0) of 2*sin(x)*cos(x)=2*0*1=0
How can you find the point at which the function is continuous?
Using calculus to see if the function f(x) is continuous at a point (point c) involves three steps. These three conditions must be met: 1. f(c) exists, is defined 2. lim f(x) exists x-->c 3. f(c)= lim f(x) x-->c
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.
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 the limit of 1-cos x over x squared when x approaches 0?
1
UP Board Intermediate Syllabus 12th?
lim x=0, cos2x-1/cosx-1
Lim x approaches 0 x x x x?
When the limit of x approaches 0 x approaches the value of x approaches infinity.
What is the limit of x divided by e to the x as x approaches negativ infinity?
lim x -> -inf [x/ex] = lim x -> +inf[-x/e-x] = - lim x -> +inf [ xex ] = -infIf you want to see this function then I suggest you use either:(a) wolframalpha.com: put in show me x/exp(x)or (b) geogebra, which is available for the desktop.
What is limit of 1 -cos x divided by x as x approaches 0?
1
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 e to the power infinity?
E to the power infinity, or lim en as n approaches infinity is infinity.
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
What does it mean when we say that function's graph approaches infinity as it gets closer to a vertical asymptote?
Definition: The line x = a is called a vertical asymptoteof the curve y = f(x) if at least one of the following statements is true:lim(x→a) f(x) = ∞; lim(x→a⁻) = ∞, lim(x→ a⁺) = ∞lim(x→a) f(x) = -∞; lim(x→a⁻) = -∞, lim(x→ a⁺) = -∞In general we write lim(x→a) f(x) = ∞ to indicate that the values of f(x) become larger and larger (or "increases without bound") as x becomes closer and closer to a. (It simply express the particular way in which the limit does not exist.)The symbol lim(x→a) f(x) = -∞ can be read as "the limit of f(x), as x approaches a, is negative infinity" or "f(x) decreases without bond as x approaches a." That is the limit does not exist.
