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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).
Wiki User
∙ 14y ago
Wiki User
∙ 14y agotan 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).
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What is the infinite limit of 1 divided by ln x?
The limit should be 0.
What is any number divided by itself?
Any number divided by itself equals 1. ex: 24 / 24 = 1 and 2,154,378,549,215,044.32158 / 2,154,378,549,215,044.32158 = 1. 0 divided by 0 is not defined, although one could define it as the limit as x approaches 0 of x/x and that is 1.
Which is true 15 divided by 0 equals 0 or 0 divided by 15 equals 0?
0 divided by15 = 0 15 divided by 0 is undetermined ( infinite)
0 divided by negative 8?
0
What is the answer to 0 divided by negative 2?
0
How do you evaluate this limit Algebraically limit as x approaches 0 of 3sinx divided by x-2tanx without using Lhospital rule?
sinx = sin0 = 0 tanx = tan0 = 0 you have 0/0 by you limit conditions
How does secx plus 1 divided by cotx equal 1 plus sinx divided by cosx?
secx = 1/cosxand 1/cotx = tanx, therefore1/cosx + tanx = 1 + sinx/cosx, andsin/cos = tanx, therefore1/cosx + tanx = 1 + tanx, therefore1/cosx = 1, therfore1 = cosx.So, therfore, it is not neccesarily true.But if you meansecx plus 1 divided by cotx equals (1 plus sinx) divided by cosx(this is probably what you mean) Let's start over!secx = 1/cosxand 1/cotx = tanx, therefore1/cosx + tanx = (1+sinx)/cosx therefore1/cosx + tanx = 1/cosx + sinx/cosxsinx/cosx = tanx therfore1/cosx + tanx = 1/cosx + tanxDo you think this is correct? Subtract both sides by 1/cosx + tanx:0 = 0So, therefore, this is correct!(BTW, I'm in Grade 6! :P)
What is the limit of x- sin x cos x over tan x -x as x tends to zero?
It is minus 1 I did this: sinx/cos x = tan x sinx x = cosx tanx you have (x - sinxcosx) / (tanx -x) (x- cos^2 x tan x)/(tanx -x) let x =0 -cos^2 x (tanx) /tanx = -cos^x -cos^2 (0) = -1
What is the derivative of 1 plus tanx?
d/dx(1+tanx)=0+sec2x=sec2x
What is the infinite limit of 1 divided by ln x?
The limit should be 0.
How do you calculate the limit of e5x -1 divided by sin x as x approaches 0?
You can use the L'hopital's rule to calculate the limit of e5x -1 divided by sin x as x approaches 0.
What is limit of 1 -cos x divided by x as x approaches 0?
1
How many times does 0 divide into 10?
0
How do you solve the limit 1 minus cos3x divided by sin3x as approaches 0?
Use l'Hospital's rule: If a fraction becomes 0/0 at the limit (which this one does), then the limit of the fraction is equal to the limit of (derivative of the numerator) / (derivative of the denominator) . In this case, that new fraction is sin(3x)/cos(3x) . That's just tan(3x), which goes quietly and nicely to zero as x ---> 0 . Can't say why l'Hospital's rule stuck with me all these years. But when it works, like on this one, you can't beat it.
What is the limit of 2x divided by tanx plus sinx as it approaches zero from the right?
This answer will assume you understand basic concepts of limits. This is what I am interpreting your problem as: lim x->0+ [(2x)/(tan(x) + sin(x))] It is easy to see that simple substitution of 0 in for x will yield an indeterminate form 0/0. So, L'Hopital's rule will be applied to solve this limit. This rule states that an indeterminate form in a limit can still be solved for by deriving the top and bottom of the divided function and resolving for the limit. The "top" of this expression is 2x, and the "bottom" is tan(x) + sin(x). Deriving both top and bottom yields a new expression: 2/(sec2(x)+cos(x)) Substitution of 0 into this expression yields a determinate form, because sec2(0)=1/cos2(0)=1/1=1 and cos(0)=1, so the new "bottom" is 1+1=2. The general limit of this new expression is equal to the general limit of the original expression, so: lim x->0 [2/(sec2(x) + cos(x))] = 2/2 = 1 = lim x->0 [(2x)/(tan(x) + sin(x))] Since this is a general limit, the limit as x approaches zero from the left and right are equal, so they are both 1. The answer is 1.
What is a limit in a number?
There is no limit in a number. There is however a mathematical concept known as a limit but this applies to functions. For instance: The limit of f(n)=(1/n) as n goes to infinity is 0. The limit of f(n)=(n/n) as n goes to infinity is 1.
What is the limit of 9x divided by 2x as x approaches 0?
9x/2x = 9/2 = 4.5
