Limit As X Approaches Infinity Of 1 2/x ^x

What is the limit as x approaches infinity of x^2-√ (x^4-x^2+2)?

A2A[math]\begin{align} x^2-\sqrt{x^4-x^2+2} &= \frac{(x^2-\sqrt{x^4-x^2+2})(x^2+\sqrt{x^4-x^2+2})}{x^2+\sqrt{x^4-x^2+2}} \\&= \frac{x^4-x^4+x^2-2}{x^2+\sqrt{x^4-x^2+2}} \\&=\frac{x^2-2}{x^2+\sqrt{x^4-x^2+2}} \\&=\frac{x^2(1-\frac{2}{x^2})}{x^2(1+\sqrt{1-\frac{1}{x^2}+\frac{2}{x^4}})} \\&=\frac{1-\frac{2}{x^2}}{1+\sqrt{1-\frac{1}{x^2}+\frac{2}{x^4}}}\end{align}[/math][math]\displaystyle\lim_{x\rightarrow\infty} 1-\frac{2}{x^2} = 1[/math][math]\displaystyle\lim_{x\rightarrow\infty} 1+\sqrt{1-\frac{1}{x^2}+\frac{2}{x^4}} = 2[/math]So[math]\displaystyle\lim_{x\rightarrow\infty} x^2-\sqrt{x^4-x^2+2} = \frac{1}{2}[/math]

What is the limit of x as it approaches infinity of x!/x^x?

It approaches [math]0[/math].Consider the Stirling’s approximation to [math]n! [/math]vis[math]\displaystyle n! \sim \sqrt[]{2\pi n }\Big(\dfrac{n}{e}\Big)^n[/math]What this basically means is that as [math]n[/math] grows rapidly and it gets really large, both the functions tend to give the same output.i.e [math]\displaystyle \lim_{n \rightarrow \infty} \frac{n!}{\sqrt[]{2\pi n} \Big(\frac{n}{e}\Big)^n} = 1[/math]Now the limit in the question is concerned with a really large value of [math]x[/math] so we can use this approximation.Thus [math]\displaystyle \lim_{n \rightarrow \infty} \frac{n!}{n^n} = \lim_{n \rightarrow \infty} \frac{n^n \sqrt[]{2\pi n}}{e^n n^n}[/math]Therefore [math]\displaystyle \lim_{n \rightarrow \infty} \frac{\sqrt[]{2\pi n}}{e^n} = \lim_{n \rightarrow \infty} \frac{\sqrt[]{2 \pi}}{2e^n \sqrt[]{n}}[/math]The above result was due to L’Hopitals rule. It allows us to solve the limits of indeterminate forms such as [math]\displaystyle( \frac{\infty}{\infty}, \frac{0}{0}, 1^{\infty}, ...)[/math] and many more by simply considering the limit of the derivatives of the numerator and denominators in the ratio.It is clear that when you substitute a really huge amount in the denominator, it will grow exponentially and as a consequence, the fraction will get closer and closer to [math]0[/math]. Thus putting it formally,[math]\displaystyle \lim_{x \rightarrow \infty} \frac{x!}{x^x} = 0[/math]

What is the limit if of xsin (1/x^2) as x approaches infinity?

To solve this, you need to put the equation in a form that allows you to use L’Hopital’s rule. Because [math]x=1/(1/x)[/math], you can rewrite [math]xsin(1/x^2)[/math] as [math]sin(x^(-2))/(x^(-1))[/math], which has limit [math]0/0[/math] as x approaches infinity. You can now apply L’Hopital’s rule, giving you [math]cos(x^(-2))*-2x^(-3)/(-x^(-2))[/math], which can be simplified to [math]2cos(x^(-2))/x[/math]. Now that you have simplified the function, you can try direct substitution again. The limit of [math]x^(-2)[/math] as x approaches infinity is 0, and the limit of [math]cos(x)[/math] as x approaches 0 is 1. We can apply those limits when we substitute infinity in for x, giving us [math]2/infinity[/math]. For any finite number x, [math]x/infinity[/math] is equal to 0, and 2 is a finite number. Therefore, the limit of [math]xsin(1/x^2)[/math] as x approaches infinity is 0.

What is the limit of [(x+2) /(x+1)] ^(x+3) as x approaches infinity?

[math]\displaystyle \lim_{x \to \infty} \left( \dfrac{x+2}{x+1} \right)^{x+3} = \displaystyle \lim_{x \to \infty} \left( 1+\dfrac{1}{x+1} \right)^{x+1} \cdot \displaystyle \lim_{x \to \infty} \left( 1+\dfrac{1}{x+1} \right)^2 [/math][math]= e \cdot 1[/math][math]=e[/math]. [math]\blacksquare[/math]

What is the limit of (1+a/x) ^bx as x approaches infinity?

User-12395599714407042467 has used the result [math]lim_{x \to \infty}(1+\frac{a}{x})^x=e^a[/math]Here is the proof for this result:Let [math](1+\frac{a}{x})^x=y[/math][math]ln(y)=ln(1+\frac{a}{x})^x=xln(1+\frac{a}{x})[/math]Rearrange the right hand side to get [math]\frac{ln(1+\frac{a}{x})}{\frac{1}{x}}[/math] so that its limit has the form [math]\frac{\infty}{\infty}[/math]Now we can apply L’Hospital’s rule to find the limit of the right hand side[math]lim_{x \to \infty}\frac{ln(1+\frac{a}{x})}{\frac{1}{x}}[/math][math]=\frac{\frac{1}{1+\frac{a}{x}} \times \frac{-a}{x^2} }{\frac{-1}{x^2}}[/math][math]lim_{x \to \infty} \frac{a}{1+\frac{a}{x}}=a[/math]Therefore, [math]y=e^a[/math]

How can the limit of (x log x) / log (x!) as x approaches infinity be proven to equal 1?

[math]L=lim_{{x}\to{\infty}} \frac{xlog(x)}{log(x!)}[/math]Deceptive pseudo-proof:Using L’Hospital’s rule, we get[math]L=lim_{{x}\to{\infty}} \frac{log(x)+1}{\frac{x!(1+1/2+1/3+1/4+....+1/x)}{x!}}[/math]Note that, x! is defined for positive integers only, hence we cannot take it’s derivative but still I have attempted to do so hence I called this as a deceptive pseudo proof.Mathematically it is illegal to take derivative of a discrete function like x!Anyways, proceeding further[math]L=lim_{{x}\to{\infty}} \frac{log(x)+1}{1+1/2+1/3+1/4+....+1/x}[/math][math]L=lim_{{x}\to{\infty}} \frac{log(x)+1}{log(x)+\gamma}[/math]Again, using L’Hospital’s rule,[math]L=lim_{{x}\to{\infty}} \frac{\frac{1}{x}}{\frac{1}{x}}[/math]After cancellation, we get[math]L=1[/math]Correct proof:We have definition of factorial as, [math]x!=x\times(x-1)\times(x-2)\times…\times3\times2\times1[/math]This can be written as,[math]x!=x\times(x-1)\times(x-2)\times....\times(x-(x-3))\times(x-(x-2))\times(x-(x-1))[/math]Taking ‘x’ common from each bracket,[math]x!=(x\times x\times x\times ......x-times)\times(1-1/x)\times(1-2/x)\times....\times(1-(x-3)/x)\times(1-(x-2)/x)\times(1-(x-1)/x)[/math][math]x!=x^x\times(1-1/x)\times(1-2/x)\times....\times(1-(x-3)/x)\times(1-(x-2)/x)\times(1-(x-1)/x)[/math]Taking log on both the sides, we get[math]log(x!)=log(x^x)+log((1-1/x)\times(1-2/x)\times....\times(1-(x-3)/x)\times(1-(x-2)/x)\times(1-(x-1)/x))[/math][math]log(x!)=xlog(x)+log((1-1/x)\times(1-2/x)\times....\times(1-(x-3)/x)\times(1-(x-2)/x)\times(1-(x-1)/x))[/math]Dividing both the sides by [math]log(x!)[/math] , we get[math]1=\frac{xlog(x)}{log(x!)}+\frac{log((1-1/x)\times(1-2/x)\times....\times(1-(x-3)/x)\times(1-(x-2)/x)\times(1-(x-1)/x))}{log(x!)}[/math]Taking limit as [math]x\to\infty[/math] , we get[math]1=lim_{{x}\to{\infty}}\frac{xlog(x)}{log(x!)}+lim_{{x}\to{\infty}}\frac{log((1-1/x)\times(1-2/x)\times....\times(1-(x-3)/x)\times(1-(x-2)/x)\times(1-(x-1)/x))}{log(x!)}[/math]Second limit here can easily be solved by directly putting limit values.So, we have[math]1=lim_{{x}\to{\infty}}\frac{xlog(x)}{log(x!)}+lim_{{x}\to{\infty}}\frac{log((1-0)\times(1-0)\times....)}{lim_{{x}\to{\infty}} log(x!)}[/math][math]1=lim_{{x}\to{\infty}}\frac{xlog(x)}{log(x!)}+\frac{log((1-0)\times(1-0)\times....)}{\infty}[/math][math]1=lim_{{x}\to{\infty}}\frac{xlog(x)}{log(x!)}+0[/math]Hence, we got [math]lim_{{x}\to{\infty}}\frac{xlog(x)}{log(x!)}=1[/math]

What is the limit as x approaches infinity of (5x^3+4) /sqrt (x^4-2)?

As x approaches infinity, you can factor out the denominator by a factor of x^4. This would become: sqrt(x^4) * sqrt(1) - sqrt(2/x^4). When x approaches infinity, sqrt(2/x^4) will become so small that we don’t consider it (limits at infinity theorem) so it would be (5x^3+4)/(sqrtx^4) which would simplify to (5x^3+4)/x^2. Now we can use limits at infinity by dividing everything by x^2. That would give us (5x + 4/x^2)/(1). Now, if we use limits at infinity theorem, as x approaches infinity, the limit would equal infinity.

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.