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Determine $\lim_{x\to\infty} f(x)$ where $$f(x)=\left(\frac{x^4+(x^2 +1)^{0.5}}{x^4-x+2}\right)^{x^3+3x}.$$

I used the binomial expansion method and my answer is $e^2$ but when I plot this function on Desmos and check value of this function at very large values like $10^9$ it shows function is equal to $1$. I am more concerned about why this happens rather than answer.

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  • $\begingroup$ Can you show your work? Ah, nm, I agree that the limit is $e^2$. $\endgroup$ – Calvin Lin Dec 14 '19 at 15:08
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    $\begingroup$ In Desmos, I'm guessing it's a calculator overflow error. $\endgroup$ – Calvin Lin Dec 14 '19 at 15:17
  • $\begingroup$ I agree with the overflow. For $x=10^9$, $\log(f(x))=2.0000000020000000065$. Try again with Desmos using the logarithm. $\endgroup$ – Claude Leibovici Dec 14 '19 at 16:27
  • $\begingroup$ I will try on other evaluation websites $\endgroup$ – chand sureja Dec 14 '19 at 16:29
  • $\begingroup$ Please show us the exact syntax you have used on Desmos. $\endgroup$ – Yves Daoust Dec 14 '19 at 16:30
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When you compute the ratio for $x=10^9$, it is very close to

$$\frac{1+10^{-27}}{1-10^{-27}}\approx1+2\cdot10^{-27}.$$

To take the power $10^{27}$ (and obtain a result very close to $e^2$), the calculator should evaluate a logarithm with at least $27$ significant digits. If it works using quadruple-precision accuracy, this would be enough to get the first few correct digits (in principle 34 decimal digits).

You could hack the internal representation of Desmos by finding the order of magnitude of $x$ that makes the value drop to $1$.


A more direct way is by evaluating expressions like

$$5-\dfrac1{\dfrac15}$$ though depending on the guard digits and rounding policy, this may return exactly $0$.

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We have that

$$\left(\frac{x^4+(x^2 +1)^{0.5}}{x^4-x+2}\right)^{x^3+3x}=\left[\left(1+\frac{(x^2 +1)^{0.5}+x-2}{x^4-x+2}\right)^{\frac{x^4-x+2}{(x^2 +1)^{0.5}+x-2}}\right]^{(x^3+3x)\frac{(x^2 +1)^{0.5}+x-2}{x^4-x+2}}$$

and

$$\left(1+\frac{(x^2 +1)^{0.5}+x-2}{x^4-x+2}\right)^{\frac{x^4-x+2}{(x^2 +1)^{0.5}+x-2}} \to e$$

then we need to evaluate

$$\lim_{x\to \infty} \frac{(x^3+3x)[(x^2 +1)^{0.5}+x-2]}{x^4-x+2}$$

which turns to be equal to $2$, therefore your result seems correct.

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  • $\begingroup$ It's not e^2 when I plug very big values it goes to 1 $\endgroup$ – chand sureja Dec 14 '19 at 16:28
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    $\begingroup$ @chandsureja. Because of the overflow. Desmos tries to compute $$1.000000000000000000000000002^{1000000000000000003000000000}$$ $\endgroup$ – Claude Leibovici Dec 14 '19 at 16:33
  • $\begingroup$ Try to take x large for the last limit. What do you obtain numerically? Maybe 2? $\endgroup$ – user Dec 14 '19 at 16:52
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Take $$\ln f(x)=({x^3+3x})\ln\left(\frac{x^4+\sqrt{x^2 +1}}{x^4-x+2}\right)$$

and note

$$\frac{x^4+\sqrt{x^2 +1}}{x^4-x+2} =\frac{1+\frac1{x^3}\sqrt{1+\frac1{x^2}}}{1-\frac1{x^3}+\frac2{x^4}} =\frac{1+\frac1{x^3}+O(\frac1{x^5})}{1-\frac1{x^3}+\frac2{x^4}} =1+ \frac2{x^3}+O(\frac1{x^4})$$

Then,

$$\lim_{x\to\infty} \ln f(x)=\lim_{x\to\infty}(x^3+3x)\ln(1+ \frac2{x^3}+O(\frac1{x^4}))$$ $$=\lim_{x\to\infty} (x^3+3x)(\frac2{x^3}+O(\frac1{x^4}))=\lim_{x\to\infty} (2+O(\frac1{x}))=2$$

Thus,

$$\lim_{x\to\infty} f(x)=e^2$$

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