Let $$f(x)$$, $$g(x)$$ and $$h(x)$$ and functions be such that $$\displaystyle\lim_{x \to{+}\infty}{f(x)}=1$$, $$\displaystyle\lim_{x \to{+}\infty}{g(x)}=-5$$ and $$\displaystyle\lim_{x \to{+}\infty}{h(x)}=-\infty$$.
Find the following limits:
a) $$\displaystyle\lim_{x \to{+}\infty}{f(x) \cdot (f(x)+g(x))}$$
b) $$\displaystyle\lim_{x \to{+}\infty}{(f(x)-g(x))^{f(x)+g(x)}}$$
c) $$\displaystyle\lim_{x \to{+}\infty}{h(x)-f(x)}$$
d) $$\displaystyle\lim_{x \to{+}\infty}{\frac{1-f(x)}{h(x)}+g(x)}$$
Development:
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) $$$\lim_{x\to{+}\infty}{[\dfrac{1-f(x)}{h(x)}+g(x)]}=\lim_{x\to{+}\infty}{[\dfrac{1-f(x)}{h(x)}]}+\lim_{x\to{+}\infty}{g(x)}=$$$ $$$=\dfrac{\lim_{x\to{+}\infty}{[1-f(x)]}}{\lim_{x\to{+}\infty}{h(x)}}+(-5)=\dfrac{1-\lim_{x\to{+}\infty}{f(x)}}{\lim_{x\to{+}\infty}{h(x)}}-5=$$$ $$$=\dfrac{1-1}{-\infty}-5=\dfrac{0}{-\infty}-5=0-5=-5$$$
Solution:
a) $$-4$$
b) $$\dfrac{1}{6^4}=\dfrac{1}{1.296}$$
c) $$-\infty$$
d) $$-5$$