Recent content by walker242

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    Simplify $ln(x(\sqrt{1+e^x}-\sqrt{e^x})) + ln(\sqrt{1+e^{-x}}+1)$

    Actually, it's possible to get further: \ln \left(x \left(\sqrt{1+e^{x}}-\sqrt{e^{x}}\right) + \ln\left(\sqrt{1+e^{-x}}+1 \right) = \ln x + \ln\left(\sqrt{1+e^{x}}-\sqrt{e^{x}}\right) + \ln\left(\sqrt{1+e^{-x}}+1\right) = = \ln x + \ln\left(\left(\sqrt{1+e^{x}}-\sqrt{e^x}\right)...
  2. W

    Limit Calculator for \sqrt{x^{2}+5} and \sqrt{x^{2}+2} with x\rightarrow \infty

    \lim_{x\rightarrow \infty} \frac{5\sqrt{x^{2}+2} + x}{2\sqrt{x^{2}+5} +x} = \lim_{x\rightarrow\infty} = \frac{5}{2}\frac{x(\sqrt{1+\frac{2}{x}})+1}{x(\sqrt{1+\frac{5}{x}})+1} = \frac{5}{2}\cdot\frac{2}{2} = \frac{5}{2} Cheers!
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    Limit Calculator for \sqrt{x^{2}+5} and \sqrt{x^{2}+2} with x\rightarrow \infty

    While I do know l'Hopitals rule, we have not yet covered it in the course. The problem should be solved without using (sadly).
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    Limit Calculator for \sqrt{x^{2}+5} and \sqrt{x^{2}+2} with x\rightarrow \infty

    Homework Statement Calculate the limit of \lim_{x\rightarrow \infty} \frac{\sqrt{x^{2}+5} - x}{\sqrt{x^{2}+2} - x} Homework Equations - The Attempt at a Solution Neither multiplying with the conjugate nor trying to break out x helps me, as I'm left with "0/0" in those cases.
  5. W

    Calculate the limit of $a_{n} = \frac{n^n}{(n-1)^n}$ when $n\rightarrow\infty$

    While both suggestions do give me the end result I want, we have still not reached MacLaurin expansions, nor L'Hopitals rule. Much thanks for your help, however.
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    Calculate the limit of $a_{n} = \frac{n^n}{(n-1)^n}$ when $n\rightarrow\infty$

    Homework Statement Calculate the limit of a_{n} = \frac{n^n}{(n-1)^n} when n\rightarrow \infty, where n is an integer. Homework Equations - The Attempt at a Solution a_{n} = \frac{n^n}{(n-1)^n} = \left(\frac{n}{n-1}\right)^{n} = e^{\ln\frac{n}{n-1}\right)^{n}} =...
  7. W

    Simplify $ln(x(\sqrt{1+e^x}-\sqrt{e^x})) + ln(\sqrt{1+e^{-x}}+1)$

    Thanks. :)
  8. W

    Simplify $ln(x(\sqrt{1+e^x}-\sqrt{e^x})) + ln(\sqrt{1+e^{-x}}+1)$

    Homework Statement As in title, simplify ln(x(\sqrt{1+e^x}-\sqrt{e^x})) + ln(\sqrt{1+e^{-x}}+1), x > 0. Homework Equations - The Attempt at a Solution So far ln(x(\sqrt{1+e^x}-\sqrt{e^x})) + ln(\sqrt{1+e^{-x}}+1) = ln(x(\sqrt{1+e^x}-\sqrt{e^x})(\sqrt{1+e^{-x}}+1)) =...
  9. W

    Conjugate Limit: $\frac{\sqrt{x}-1}{x-1}$

    \lim_{x\to 1} \frac{\sqrt{x}-1}{x-1} = \lim_{x\to 1} \frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1)}{\left(x-1\right)\left(\sqrt{x}+1\right)} = \lim_{x\to 1} \frac{x-1}{x\sqrt{x}+x-\sqrt{x}-1} = \lim_{x\to1}\frac{x-1}{\sqrt{x}\left(x-1\right)+x-1} = \lim_{x\to1}\frac{x-1}{(x-1)(\sqrt{x}+1)} =...
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    Conjugate Limit: $\frac{\sqrt{x}-1}{x-1}$

    $\lim_{x\to 1}\frac{\sqrt{x}-1}{x-1}$ Homework Statement Calculate the limit of \lim_{x\to 1}\frac{\sqrt{x}-1}{x-1}. Homework Equations As above. The Attempt at a Solution Have tried to multiplicate with the conjugate.
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    Prove that 3n^2 - 1 can't be a square of a integer n

    Thanks for the reply! While that probably is one way of looking at the problem, we haven't yet reached modulus in our studies.
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    Prove that 3n^2 - 1 can't be a square of a integer n

    Well, the problem statement is in the title: Given that n is an integer, show that 3n2 - 1 can't be the square of an integer. Currently, I don't have any idea at all where to start. Method is probably to assume opposite and show that this leads to a contradiction. Any hint as to where to...
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