Finite and infinite limits

Using the properties of the finite and infinite limits, we will find the following limits:

a) $$\displaystyle\lim_{x \to{+}\infty}{[f(x)\cdot(f(x)+g(x))]}=\lim_{x \to{+}\infty}{f(x)}\cdot\lim_{x \to{+}\infty}{[f(x)+g(x)]}=$$ $$=1\cdot[\lim_{x \to{+}\infty}{f(x)}+\lim_{x \to{+}\infty}{g(x)} ]=1\cdot(1-5)=-4$$

b) Since $\displaystyle\lim_{x\to{+}\infty}{[f(x)-g(x)]}=\displaystyle\lim_{x\to{+}\infty}{f(x)}-\displaystyle\lim_{x\to{+}\infty}{g(x)}=1-(-5)=6 > 0$, we can apply the following property: $$\displaystyle\lim_{x\to{+}\infty}{[(f(x)-g(x))^{f(x)+g(x)}]}=[\displaystyle\lim_{x\to{+}\infty}{[f(x)-g(x)]}]^{\displaystyle\lim_{x\to{+}\infty}{[f(x)+g(x)]}}=$$ $$=[\lim_{x\to{+}\infty}{f(x)}-\lim{x\to{+}\infty}{g(x)}]^{\lim_{x\to{+}\infty}{f(x)}+\lim_{x\to{+}\infty}{g(x)}}=$$ $$=(1-(-5))^{(1+(-5))}=6^{-4}=\dfrac{1}{6^4}=\dfrac{1}{1.296}$$

c) $$\lim_{x\to{+}\infty}{[h(x)-f(x)]}=\lim_{x\to{+}\infty}{h(x)}-\lim_{x\to{+}\infty}{f(x)}=-\infty-1=-\infty$$

d) $$\lim_{x\to{+}\infty}{[\dfrac{1-f(x)}{h(x)}+g(x)]}=\lim_{x\to{+}\infty}{[\dfrac{1-f(x)}{h(x)}]}+\lim_{x\to{+}\infty}{g(x)}=$$ $$=\dfrac{\lim_{x\to{+}\infty}{[1-f(x)]}}{\lim_{x\to{+}\infty}{h(x)}}+(-5)=\dfrac{1-\lim_{x\to{+}\infty}{f(x)}}{\lim_{x\to{+}\infty}{h(x)}}-5=$$ $$=\dfrac{1-1}{-\infty}-5=\dfrac{0}{-\infty}-5=0-5=-5$$