The limit as x approaches infinity of (36x^2+x)^(1/2)-6x?
lim √36x²+x -6x x->∞ lim [√36x²+x -6x]*[√36x²+x +6x]/[√36x²+x +6x] x->∞ lim [36x²+x -36x²]/[√x²(36+1/x) +6x] x->∞ lim x/x[√(36 +1/x)+6] x->∞ lim 1/[√(36 +1/x)+6] x->∞ as lim x->∞ then 1/[√(36 +1/∞)+6] =>1/[6+6] =1/12 the limit is 1/12
What is the limit of sinx as x approaches infinity?
No. The correct answer is that FIRST you establish whether a value has a limit, and THEN you may calculate it.The definition of a limit is, as my feeble old mind recalls it, that for any restraint (fluctuation) on the result you desire, you may find a matching restraint on the input (x really large), so the result will stay within bounds.So here goes (this is tough): IF you can restrain the input SO THAT the result is restrained THEN you have a limit value, OTHERWISE you do not have a limit value.Here, the result will fluctuate in [-1, 1] no matter how large x is, so there is no limit value.Here’s a real limit that does exist: What’s the limit on x/2 as x goes to 4? If you need to restrain the answer to 0.5, then you can restrain x to [3, 5], and the answer will be in [1.5, 2.5], and you can further restrain the answer by restraining the input.
Explain how limit as x approaches infinity of (sqrt(x^2+4x+1)-x)=2? Question is attached in a screenshot number 17.?
sqrt(x^2 + 4x + 1) - x => (sqrt(x^2 + 4x + 1) - x) * (sqrt(x^2 + 4x + 1) + x) / (sqrt(x^2 + 4x + 1) + x) => (x^2 + 4x + 1 - x^2) / (sqrt(x^2 + 4x + 1) + x) => (4x + 1) / (sqrt(x^2 + 4x + 1) + x) => (4x + 1) / (sqrt(x^2 * (1 + 4/x + 1/x^2)) + x) => (4x + 1) / (sqrt(x^2) * sqrt(1 + 4/x + 1/x^2) + x) => (4x + 1) / (x * sqrt(1 + 4/x + 1/x^2) + x) => (4x + 1) / (x * (sqrt(1 + 4/x + 1/x^2) + 1)) => x * (4 + 1/x) / (x * (sqrt(1 + 4/x + 1/x^2) + 1)) => (4 + 1/x) / (sqrt(1 + 4/x + 1/x^2) + 1) Now we can apply our limit. x goes to infinity (4 + 1/inf) / (sqrt(1 + 4/inf + 1/inf^2) + 1) => (4 + 0) / (sqrt(1 + 0 + 0) + 1) => 4 / (sqrt(1) + 1) => 4 / (1 + 1) => 4/2 => 2
Limit as x approaches infinity?
Hi You can multiply by the conjugate to get rid of the square root in the numerator. lim (x->∞) [√(64x^2 + x) - 8x] = lim (x->∞) [√(64x^2 + x) - 8x][√(64x^2 + x) + 8x]/[√(64x^2 + x) + 8x] = lim (x->∞) [64x^2 + x - 64x^2]/[√(64x^2 + x) + 8x] = lim (x->∞) x/[√(64x^2 + x) + 8x] Now, you can factor out an x from the denominator to cancel out with the x in the numerator. lim (x->∞) x/[√(64x^2 + x) + 8x] = lim (x->∞) x/[√[(x^2)(64 + 1/x)] + 8x] = lim (x->∞) x/[x√(64 + 1/x) + 8x] = lim (x->∞) x/[x(√(64 + 1/x) + 8)] = lim (x->∞) 1/[√(64 + 1/x) + 8] Now, you can use direct substitution. 1/x approaches 0 as x approaches infinity. 1/[√(64 + 0) + 8] = 1/16 I hope this helps!
What is the limit of f(x) =x-2+sqrt (x^2-8x+15) as x approaches -infinity?
Although not needed, I prefer to let x → -x so as to have the equivalent limit as x approaches infinity of -x + sqr(x^2 + 8x + 15) -2. Lets forget about the ending -2 for now. We are thus concerned with -x + sqr(x^2 + 8x + 15). Multiply & divide this by (-x - sqr(x^2 + 8x + 15)). Simplifying the top & bottom then gives-(8x + 15) /-(x + sqr(x^2 + 8x + 15)).Factoring out x^2 from the sqr and then factoring out x from the bottom & simplifying a little gives(8x + 15) / ( x * ( 1 + sqr(1 + 8/x + 15/x²))). Splitting & simplifying gives8/(1 + sqr(1 + 8/x + 15/x²)) + 15/( x * ( 1 + sqr(1 + 8/x + 15/x²))).Now, each of these can easily be evaluated at infinity. The first expression gives 8/2 = 4 and the other expression gives 0. Convince yourself of these. So the limit is 4. But not forgetting our “-2” in the original question, your sought limit is thus 4–2 = 2.What to remember? Often for these type of limits [??? + sqr(???)], we multiply the top & bottom by the “conjugate” as we did. Then its just a bunch of algebra to simplify and the answer appears.
The limit as x approaches infinity of sqrt(2x^2+1)/(3x-5)?
I have the solution, but there is something I dont understand of it. They divide the numerator and the denominator by x^2 (that I get why) but when dividing the numerator they say it is sqrt(2x^2+1)=sqrt(2+1/x^2) which doesn't make sense because you cannot say that sqrt(a+b)/c is equal to sqrt(a/c + b/c) so how do you explain the above? Also, on my book there's a side note that says "(since sqrt(x^2)=x when x>0)" but I dont see how to explains anything.
Limit as t approaches +infinity (-5t-9)/(sqrt(t^2-6t+5). Help and Please explain. Thanks in advance?
we divide top and bottom of the equation by the greatest power of 't' . since the bottom is a square root, we divide the bottom part of the equation by square root of t^2. we divide the top part by simply t-since t and square root of t^2 are the same.so we get: -5 -(9/t) --------------------------- {1-(6/t)+(5/t^2) }^1/2 now we plug in infinity for t- and so we get -5-0 ------------------------- {1-0+0}^1/2 therefore the answer is -5
Limit as x approaches negative infinity (sqrt(9x^6 - x))/(x^3+1) = ? The answer is -3 but I don't understand?
So the way I did it, since you get an indeterminate form of +∞/-∞ when you solve directly is that I divided the numerator and denominator by the highest exponent power of x (which is x³). So I got: √(9x^6 - x) ∙ (1/x³) -------------------------- (x³ + 1) ∙ (1/x³) But since the numerator is under a square root sign, to simplify, you need to change the x³ into a square root, too. You might think you can just substitute x³ = √(x^6), but for negative numbers of x (which is what you will have, since you are taking the limit when x --> -∞) x³ will be a negative number. In other words: x³ = -√(x^6) ≠ √(x^6) So you get: √(9x^6 - x) ∙ (-1/√(x^6)) -------------------------- (x³ + 1) ∙ (1/x³) This simplifies to: -√(9 - (1/x^5)) -------------------- 1 + 1/x³ Since 1/x^5 and 1/x³ both go to zero as x goes to negative infinity, you get: -√(9 - 0) ------------ 1 + 0 = -3 I hope that helps!
What is the limit of Sin (π/x) as x approaches 0?
The short answer is that your limit doesn’t exist.Some details below.Define[math]\displaystyle f(x)=\sin{\frac{\pi}{x}}[/math]Since we are interested in the behaviour as [math]x\to\,0[/math] we investigate (equivalently)[math]\displaystyle f(\frac{1}{k})[/math] as [math]k\to\pm\infty[/math]where k takes one of the two forms[math]n[/math] (an integer)[math]n+\frac{1}{2}[/math]Observe that[math]f(\frac{1}{k})=\sin{k\,\pi}[/math]Now let’s examine particular values[math]\begin{matrix}\textrm{x} & \textrm{f(x)} \\ \frac{1}{10} &0 \\ \frac{1}{10.5}&1 \\ \frac{1}{11}&0 \\ \frac{1}{11.5} &-1 \\ \frac{1}{1000000}&0\\ \frac{1}{1000000.5}&1 \\ \frac{1}{1000001} & 0 \\\frac{1}{1000001.5} & -1 \\\frac{1}{1,000,000,000,000,000.5} &1\\\frac{1}{1,000,000,000,000,000,001.5} & -1\end{matrix}[/math]So, in particular, very small changes of x near 0 create big changes in [math]f(x)[/math]Therefore, the limit you requested does not exist.