lim (x→0) [(x - sin x)/(tan x - x)]
Since both the numerator and the denominator have limit zero as x tends to 0, the quotient is indeterminate at 0 and of the form 0/0. Therefore, we apply the l'Hopital's Rule and the limit equals
lim (x→0) [(x - sin x)'/(tan x - x)']
= lim (x→0) [(1 - cos x)/(sec2 x - 1)] (form 0/0, use again the l'Hopital's Rule)
= lim (x→0) [(1 - cos x)'/(sec2 x - 1)']
= lim (x→0) [(0 - (-sin x)/(2sec x sec x tan x - 0)]
= lim (x→0) [(sin x)/(2sec2 x tan x)] (substitute 1/cos2 x for sec2 x and sin x/cos x for tan x)
= lim (x→0) [(sin x)/(2sin x/cos3 x)]
= lim (x→0) [(sin x cos3 x)/2sin x]
= lim (x→0) (cos3 x/2)
= 1/2
Thus, (x - sin x)/(tan x - x) tends to 0.5 as x tends to 0.
1
First to simplify matters, change y=9x. So we are looking at limit sin(y) divided by tan(y).Now lets look at right angled triangle wheresin(y) = a/ctan(y) = a/bthus we are looking at the limit of (a/c)/(a/b) = limit of b/cAs the angle y shrinks, the right angle remains constant, and the remaining angle approaches a right angle. Thus at the limit we have a triangle with equal angles and thus where b=c.As a result limit you are trying to calculate is 1.
lim(h→0) (sin x cos h + cos x sin h - sin x)/h As h tends to 0, both the numerator and the denominator have limit zero. Thus, the quotient is indeterminate at 0 and of the form 0/0. Therefore, we apply l'Hopital's Rule and the limit equals: lim(h→0) (sin x cos h + cos x sin h - sin x)/h = lim(h→0) (sin x cos h + cos x sin h - sin x)'/h' = lim(h→0) [[(cos x)(cos h) + (sin x)(-sin h)] + [(-sin x)(sin h) + (cos x)(cos h)] - cos x]]/0 = cosx/0 = ∞
"First principles" in this context means that you:* Calculate the value of the function, at some point "x+h" * Calculate the value of the function, at some point "x" * Subtract the first result minus the second result * Divide all this by "h" * See what happens when you make "h" smaller and smaller (when it tends to zero) As a formula: F(x)' = lim (as h --> 0) [F(x+h) - F(x)] / h
It is called the line of best fit because it tends to satisfy all the possible points in consideration at the same time with minimal variation.
1
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.
Since x is not a part of the expression, x can approach zero without any effect. So, the answer would be (tank-sink whole)/k, k<>0.
First to simplify matters, change y=9x. So we are looking at limit sin(y) divided by tan(y).Now lets look at right angled triangle wheresin(y) = a/ctan(y) = a/bthus we are looking at the limit of (a/c)/(a/b) = limit of b/cAs the angle y shrinks, the right angle remains constant, and the remaining angle approaches a right angle. Thus at the limit we have a triangle with equal angles and thus where b=c.As a result limit you are trying to calculate is 1.
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.
the answer is infinte.reason:when you add the two terms ,the numerator is a quadratic equation but denominator is a linear in x ,,,, so as x tends to infinite the function tends to infinite
1
No. If x tends to infinite, 1/x tends to zero.
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.
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
limit x tends to infinitive ((e^x)-1)/(x)
Unfortunately, the browser used by Answers.com for posting questions is incapable of accepting mathematical symbols. This means that we cannot see the mathematically critical parts of the question. We are, therefore unable to determine what exactly the question is about and so cannot give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "times", "divided by", "equals" etc. As it appears, you seem to be seeking the limit of sin(4x)*sin(6x) as x tends to 0. Both components of the product tend to 0 as x tens to 0 and so the limit is 0. Bit I suspect that is not the limit that you are looking for.