Find the following limits: a) $$\displaystyle\lim_{x \to{+}\infty}{\frac{\ln(x) \cdot (x^3-x^4)}{x-2x^5+3x^3}}$$ b) $$\displaystyle\lim_{x \to{+}\infty}{\Big(\frac{2x-x^2}{1+x}-x\Big)}$$ c) $$\displaystyle\lim_{x \to{+}\infty}{\Big(\frac{1+x}{x}-\frac{1}{x}+1\Big)^{\ln(x)}}$$
Development:
a) $$$\lim_{x \to{+}\infty}{\dfrac{\ln(x) \cdot (x^3-x^4)}{x-2x^5+3x^3}}=\lim_{x \to{+}\infty}{\dfrac{-x^4\cdot\ln(x)}{-2x^5}}=\lim_{x \to{+}\infty}{\dfrac{-\ln(x)}{-2x} }=0$$$
b) $$$\lim_{x \to{+}\infty}{\Big[\dfrac{2x-x^2}{1+x}-x\Big]}=\lim_{x \to{+}\infty}{\Big[\dfrac{2x-x^2}{1+x}-\dfrac{x(1+x)}{1+x}\Big]}=$$$ $$$=\lim_{x \to{+}\infty}{\Big[\dfrac{2x-x^2-x-x^2}{1+x}\Big]}=\lim_{x \to{+}\infty}{\dfrac{-2x^2}{x}}=-\infty$$$
c) $$$\lim_{x \to{+}\infty}{\Big[\dfrac{1+x}{x}-\dfrac{1}{x}\Big]^{\ln(x)}}=\lim_{x \to{+}\infty}{\Big[\dfrac{1+x-1}{x}\Big]^{\ln(x)}}=$$$ $$$=\lim_{x \to{+}\infty}{1^{\ln(x)}}=\lim_{x \to{+}\infty}{1}=1$$$
Solution:
a) $$0$$
b) $$-\infty$$
c) $$1$$$