Find the following limits:
a) $$\displaystyle\lim_{x \to{+}\infty}{\Big[\dfrac{x^2+x-1}{x^2+2}\Big]^{3x-1}}$$
b) $$\displaystyle\lim_{x \to{+}\infty}{\Big[1+\dfrac{3}{2x}\Big]^{5x}}$$
Development:
a) $$$\lim_{x\to{+}\infty}{\Big[\dfrac{x^2+x-1}{x^2+2}\Big]^{3x-1}}=1^{+\infty} \Rightarrow e^{\lim_{x\to{+}\infty}{\Big[\frac{x^2+x-1}{x^2+2}\Big](3x-1)}}=$$$ $$$e^{\lim_{x\to{+}\infty}{\Big[\frac{x-3}{x^2+2}\Big](3x-1)}}=e^{\lim_{x\to{+}\infty}{\frac{3x^2-10x+3}{x^2+2}}}=e^3$$$
b) $$$\lim_{x\to{+}\infty}{\Big[1+\dfrac{3}{2x}\Big]^{5x}}=\lim_{x\to{+}\infty}{\Big[1-\dfrac{3}{2x}\Big]^{-5x}}=$$$ $$$e^{\lim_{x\to{+}\infty}{\Big[1-\frac{3}{2x}+1\Big]^{(-5x)}}}=e^{\lim_{x\to{+}\infty}{\frac{3}{2x}5x}}=e^{\frac{3\cdot5}{2}}=e^{\frac{15}{2}}$$$
Solution:
a) $$e^3$$
b) $$e^{\frac{15}{2}}$$