Răspuns:
Explicație pas cu pas:
[tex]ex1.~ \lim_{x \to 1} \frac{f(x)}{x-1} =(\frac{0}{0})= \lim_{x \to 1} [f(x)*\frac{1}{x-1} ] = \lim_{x \to 1} [\frac{x^{2}-5x+4}{x} *\frac{1}{x-1}] = \lim_{x \to 1} \frac{(x-1)(x-4)}{x(x-1)}= \lim_{x \to 1} \frac{x-4}{x}=\frac{1-4}{1} =-3.\\ex2.~a)~f(7-0)=\sqrt{3*7+4}=\sqrt{25}=5;~~f(7)= \sqrt{3*7+4}=\sqrt{25}=5;~~f(7+0)=2*7-9=14-9=5.\\ Deoarece~f(7-0)=f(7)=f(7+0)~rezulta~ca~f(x)~este~continua~in~x=7.\\b)~f(7)-f(10)=???~~f(7)=5;~f(10)=2*10-9=11,~deci~f(7)-f(10)=5-11=-6.\\ex3.~f(x)=\frac{2x-1}{x^{2}}[/tex]
[tex]f(2x)=\frac{2*2x-1}{(2x)^{2}},~\frac{f(x)}{f(2x)}=f(x):f(2x)=\frac{2x-1}{x^{2}}:\frac{2*2x-1}{(2x)^{2}}=\frac{2x-1}{x^{2}}*\frac{4x^{2}}{4x-1}=\frac{8x-4}{4x-1},~deci\\ \lim_{x \to \infty} \frac{f(x)}{f(2x)}= \lim_{x \to \infty}\frac{8x-4}{4x-1}= \lim_{x \to \infty} \frac{x(8-\frac{4}{x} }{x(4-\frac{1}{x} }=\frac{8-0}{4-0} =2.\\\\Ex4.~ \lim_{x \to 1} \frac{1-x^{2}}{f(x)}=(\frac{0}{0})=\lim_{x \to 1} \frac{1-x^{2}}{x(1-x)}=\lim_{x \to 1}\frac{(1-x)(1+x)}{x(1-x)} =\lim_{x \to 1}\frac{1+x}{x}[/tex]
=(1+1)/1=2/1=2.
