Problems from Finite and infinite limits

Development:

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

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

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

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

d) $$\underset{x\to +\mathrm{\infty }}{lim}[{\displaystyle \frac{1-f(x)}{h(x)}}+g(x)]=\underset{x\to +\mathrm{\infty }}{lim}[{\displaystyle \frac{1-f(x)}{h(x)}}]+\underset{x\to +\mathrm{\infty }}{lim}g(x)=$$ $$={\displaystyle \frac{\underset{x\to +\mathrm{\infty }}{lim}[1-f(x)]}{\underset{x\to +\mathrm{\infty }}{lim}h(x)}}+(-5)={\displaystyle \frac{1-\underset{x\to +\mathrm{\infty }}{lim}f(x)}{\underset{x\to +\mathrm{\infty }}{lim}h(x)}}-5=$$ $$={\displaystyle \frac{1-1}{-\mathrm{\infty }}}-5={\displaystyle \frac{0}{-\mathrm{\infty }}}-5=0-5=-5$$

Solution:

a) $-4$

b) ${\displaystyle \frac{1}{6^4}}={\displaystyle \frac{1}{1.296}}$

c) $-\mathrm{\infty }$

d) $-5$

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