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 $e^4$.
Answer
$$\ln\left(\lim_{x\to +\infty}\left(\frac{x^2+5x+3}{x^2+x+3}\right)^x\right)=\lim_{x\to +\infty}\left(x(\ln(x^2+5x+3)-\ln(x^2+x+3))\right)$$
$$=\lim_{x\to +\infty}\frac{\ln(x^2+5x+3)-\ln(x^2+x+3)}{\frac{1}{x}}$$
$$\stackrel{\text{L'Hop}}=\lim_{x\to +\infty}\frac{\frac{2x+5}{x^2+5x+3}-\frac{2x+1}{x^2+x+3}}{-\frac{1}{x^2}}$$
$$=\lim_{x\to +\infty}\left(\frac{x^2(2x+1)}{x^2+x+3}-\frac{x^2(2x+5)}{x^2+5x+3}\right)$$
$$=\lim_{x\to +\infty}\frac{4x^2(x^2-3)}{(x^2+x+3)(x^2+5x+3)}$$
$$=\lim_{x\to +\infty}\frac{4(1-\frac{3}{x^2})}{(1+\frac{1}{x}+\frac{3}{x^2})(1+\frac{5}{x}+\frac{3}{x^2})}=4$$
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