# Limit of $\left(\frac{x^2+5x+3}{x^2+x+3}\right)^x$

We have to evaluate:
$$\lim\limits_{x\to\infty} \left(\frac{x^2+5x+3}{x^2+x+3}\right)^x$$

My work:

Let the desired limit equal a constant $L$.

When I take $\log$ of both sides, the exponent $x$ comes down. What do I do now? Where will we apply L'Hopital's rule? Can we do it without the rule also?

The answer is $e^4$.

• In your next step, move the $x$ to the denominator as $1/x$ to make a $0/0$ fraction and use L'Hospital's. – T.J. Gaffney Jun 2 '15 at 16:24
• It becomes too much complicated? And still no success. – Aditya Agarwal Jun 2 '15 at 16:29

$$\ln\left(\lim_{x\to +\infty}\left(\frac{x^2+5x+3}{x^2+x+3}\right)^x\right)=\lim_{x\to +\infty}\left(x(\ln(x^2+5x+3)-\ln(x^2+x+3))\right)$$

$$=\lim_{x\to +\infty}\frac{\ln(x^2+5x+3)-\ln(x^2+x+3)}{\frac{1}{x}}$$

$$\stackrel{\text{L'Hop}}=\lim_{x\to +\infty}\frac{\frac{2x+5}{x^2+5x+3}-\frac{2x+1}{x^2+x+3}}{-\frac{1}{x^2}}$$

$$=\lim_{x\to +\infty}\left(\frac{x^2(2x+1)}{x^2+x+3}-\frac{x^2(2x+5)}{x^2+5x+3}\right)$$

$$=\lim_{x\to +\infty}\frac{4x^2(x^2-3)}{(x^2+x+3)(x^2+5x+3)}$$

$$=\lim_{x\to +\infty}\frac{4(1-\frac{3}{x^2})}{(1+\frac{1}{x}+\frac{3}{x^2})(1+\frac{5}{x}+\frac{3}{x^2})}=4$$

• But the answer is $e^4$ – Aditya Agarwal Jun 2 '15 at 16:36
• @AdityaAgarwal The final number is the natural logarithm of your limit. – egreg Jun 2 '15 at 16:38
• Just the answer I wanted! – Aditya Agarwal Jun 2 '15 at 16:43

You can do so: $$\frac{x^2+5x+3}{x^2+x+3} = 1 + \frac{4x}{x^2+x+3}\sim 1 + \frac 4x$$ at $x=\infty$ and therefore $$\lim_{x\to\infty}\left(\frac{x^2+5x+3}{x^2+x+3}\right)^{\!x} = \lim_{x\to\infty} \left(1+\frac 4x\right)^{\!x} = e^4$$

• Can you plz give a more rigorous proof? – Aditya Agarwal Jun 2 '15 at 16:33
• @AdityaAgarwal It is rigorous! – egreg Jun 2 '15 at 16:38
• You forgot the power $x$ in the first limit of the last line. – anderstood Jun 2 '15 at 16:45
• @anderstood, thanks – Michael Galuza Jun 2 '15 at 16:46
• As it is not the case in general that $f(x)\sim g(x)$ implies $f(x)^x\sim g(x)^x$, to make this rigorous requires more justification of the second to last equals sign, doesn't it? @egreg – Jonas Meyer Jun 2 '15 at 18:10

$$\left[\lim_{x\to\infty}\left(1+\dfrac{4x}{x^2+x+3}\right)^{\dfrac{x^2+x+3}{4x}}\right]^{\lim_{x\to\infty}\dfrac{4x^2}{x^2+x+3}}$$

Set $\dfrac{x^2+x+3}{4x}=n$ in the inner limit to find $=e$ as $x\to\infty,n\to\infty$

and $\lim_{x\to\infty}\dfrac{4x^2}{x^2+x+3}=\lim_{x\to\infty}\dfrac4{1+\dfrac1x+\dfrac3{x^2}}=\cdots$

look at how $\frac{x^2 + 5x + 3}{x^2 + x + 3}$ behaves for large $x.$

we have $$\frac{x^2 + 5x + 3}{x^2 + x + 3} = 1 + \frac{4x}{x^2 + x+3} = 1 + 4x(x^2 + x+3)^{-1} = 1 + 4x\left(x^{-2} -x^{-4}(x+3) + \cdots \right) = 1 + \frac4x + \cdots$$

therefore $$\left(\frac{x^2 + 5x + 3}{x^2 + x + 3} \right)^x = \left(1 + \frac4x+\cdots\right)^x = e^4 \text{ as } x \to \infty.$$

• How did you change $(x^2+x+3)^{-1}$ to that? And how did you deduce that in the end? – Aditya Agarwal Jun 2 '15 at 16:35
• @AdityaAgarwal., $\frac1{x^2 + x+3} = (x^2 + x + 3)^{-1}.$ then use the binomial theorem $(big + small)^{-1} = big^{-1} - big^{-2}\, small + \cdots$ – abel Jun 2 '15 at 16:36
• Nice use of expansion. The way I did it in my head was lazy division of $x$. Nearly hitting the 20k mark :). – Chinny84 Jun 2 '15 at 16:37
• @Chinny84, thanks. good to see you active again. seen any nice nonlinear differential equation? – abel Jun 2 '15 at 16:38
• Sadly not so much. not many exciting ones (just the multiply by $\dot{y}$ and reduce :(. I am working on my own nonlinear ode from my thesis (I would share it on here with more accomplished mathematicians such as yourself, but I am being selfish ;) ). – Chinny84 Jun 2 '15 at 16:41