Solution verification: $\lim_{x\to \infty}\Bigg(\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}\Bigg)^{x\ln x}$

Question

Find: $$\displaystyle\lim_{x\to \infty}\left(\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}\right)^{x\ln x}$$

My attempt:

$\displaystyle\lim_{x\to\infty}\left(\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}\right)^{x\ln x}=\lim_{x\to\infty}\left(1+\frac{\ln(x^2+3x+4)-\ln(x^2+2x+3)}{\ln(x^2+2x+3)}\right)^{x\ln x}=\\\displaystyle\lim_{x\to\infty}\left(1+\frac{\ln\left(\frac{x^2+3x+4}{x^2+2x+3}\right)}{\ln(x^2+2x+3)}\right)^{x\ln x}=\lim_{x\to\infty}\left(1+\frac{\ln\left(1+\frac{x+1}{x^2+2x+3}\right)}{\ln(x^2+2x+3)}\right)^{x\ln x}$

What I used: $\displaystyle\lim_{x\to\infty}\ln\left(1+\frac{x+1}{x^2+2x+3}\right)=0\;\;\&\;\;\lim_{x\to\infty}\ln(x^2+2x+3)=+\infty$

In the end, I got an indeterminate form: $\displaystyle\lim_{x\to\infty}1^{x\ln x}=1^{\infty}$

Have I made a mistake anywhere? It seems suspicious.

added: replacement $\frac{x+1}{x^2+2x+3}$ by $\frac{1}{x}$ wasn't appealing either.

Would:$$\lim_{x\to\infty}\Big(\Big(1+\frac{1}{x}\Big)^x\Big)^{\ln x}=x=\infty$$ be wrong?

//a few days after users had provided hints and answered the question,we discussed this with our assistant and he suggested a table formula that can also be applied (essentialy the last step in methods provided in the answers I recieved).:$$\lim_{x\to c}f(x)=1\;\&\;\lim_{x\to c}g(x)=\pm\infty$$then$$\lim_{x\to c}f(x)^{g(x)}=e^{\lim_{x\to c}(f(x)-1)g(x)}//$$

Answer 1

By your work and since $e^x$ and $\ln$ are continuous function, we obtain: $$\lim_{x\rightarrow+\infty}\left(\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}\right)^{x\ln{x}}=\lim_{x\rightarrow+\infty}\left(1+\frac{\ln\frac{x^2+3x+4}{x^2+2x+3}}{\ln(x^2+2x+3)}\right)^{\frac{\ln(x^2+2x+3)}{\ln\frac{x^2+3x+4}{x^2+2x+3}}\cdot\frac{x\ln{x}\ln\frac{x^2+3x+4}{x^2+2x+3}}{\ln(x^2+2x+3)}}=$$ $$=e^{\lim\limits_{x\rightarrow+\infty}\frac{\ln{x}\ln\left(1+\frac{x+1}{x^2+2x+3}\right)^x}{2\ln{x}+\ln\left(1+\frac{2}{x}+\frac{3}{x^2}\right)}}=e^{\frac{1}{2}\ln\lim\limits_{x\rightarrow+\infty}\left(1+\frac{x+1}{x^2+2x+3}\right)^{\frac{x^2+2x+3}{x+1}\cdot\frac{x(x+1)}{x^2+2x+3}}}=e^{\frac{1}{2}\ln{e}}=\sqrt{e}.$$

Answer 2

Use $\lim_{t\to 1}\frac{\ln(t)}{t-1}=1$ twice: first for $t=\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}$ and then for $t=\frac{x^2+3x+4}{x^2+2x+3}$. The rest is simplification. $$\begin{align} \lim_{x\to\infty}\left(\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}\right)^{x\ln x} &= \exp\left(\lim_{x\to\infty}\ln\left(\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}\right)x\ln x\right) \\ &= \exp\left(\lim_{x\to\infty}\frac{\ln\frac{x^2+3x+4}{x^2+2x+3}}{\ln(x^2+2x+3)}x\ln x\right) \\ &=\exp\left(\lim_{x\to\infty}\frac{(x+1)x}{x^2+2x+3}\frac{\ln x}{\ln(x^2+2x+3)}\right) \end{align} $$ We have $$\frac{(x+1)x}{x^2+2x+3}=\frac{x^2+x}{x^2+2x+3}\to 1 $$ For the other fraction, use that $\ln(x^2+2x+3)=\ln(x^2(1+2/x+3/x^2))=2\ln x+\ln(1+2/x+3/x^2)$ $$\frac{\ln x}{\ln(x^2+2x+3)}=\frac{\ln x}{2\ln x+\ln(1+2/x+3/x^2)}=\frac{1}{2+\frac{\ln(1+2/x+3/x^2)}{\ln x}}\to \frac 12 $$ So the limit is $e^{\frac 12 \cdot 1}=\sqrt{e}$.

Answer 3

Whenever you see an expression of type $\{u(x) \} ^{v(x)} $ under limit it is best to take logarithms. This makes your expressions simpler and easier to type and write thereby allowing you to focus on the problem more efficiently.

Thus if $L$ is the desired limit then \begin{align} \log L&=\lim_{x\to\infty} (x\log x)\log\frac{\log(x^2+3x+4)}{\log(x^2+2x+3)}\notag\\ &=\lim_{x\to\infty} (x\log x) \log\left(1+\dfrac{\log\dfrac{x^2+3x+4}{x^2+2x+3}}{\log(x^2+2x+3)}\right)\notag\\ &=\lim_{x\to \infty} (x\log x) \frac{\log(1+f(x))}{f(x)}\cdot f(x)\notag\\ &=\lim_{x\to \infty} x\log x\cdot 1\cdot f(x) \notag\\ &=\lim_{x\to\infty} x\log x\cdot\dfrac{\log\dfrac{x^2+3x+4}{x^2+2x+3}} {\log(x^2+2x+3)} \notag\\ &=\lim_{x\to\infty} \frac{x} {2}\log\left(1+\frac{x+1}{x^2+2x+3}\right)\frac{\log x^2}{\log(x^2+2x+3)}\notag\\ &=\lim_{x\to\infty} \frac{x} {2}\frac{\log(1+g(x))}{g(x)}\cdot g(x) \left(1+\dfrac{\log\dfrac{x^2}{x^2+2x+3}}{\log(x^2+2x+3)}\right)\notag\\ &=\lim_{x\to \infty} \frac{x} {2}\cdot 1\cdot g(x)\cdot 1 \notag\\ &=\lim_{x\to\infty} \frac{x(x+1)}{2(x^2+2x+3)}\notag\\ &=\frac{1}{2}\notag \end{align} and hence $L=\sqrt{e}$. You should be able to notice that both $$f(x)=\dfrac{\log\dfrac{x^2+3x+4}{x^2+2x+3}}{\log(x^2+2x+3)} , g(x)=\frac{x+1}{x^2+2x+3} $$ clearly tend to $0$ as $x\to \infty $.