- #1
John O' Meara
- 330
- 0
Find the smallest positive number N such that for each x in the interval (N,+infinity) the value of [tex] f(x)= \frac{x}{x+1} [/tex] is within .01 units of L=1.
From the definition [tex]|f(x)-L|<\epsilon \mbox{ if } x > N, \epsilon>0, N>0[/tex]. It follows that [tex] |\frac{x}{x+1}-1| < \epsilon \mbox{ if } x > N \mbox{ therefore } |\frac{-1}{x+1}| < \epsilon \mbox{ if } x > N[/tex] Now if I move the minus sign across to epsilon I get [tex] |\frac{1}{x+1}| > -\epsilon \mbox{ if } x > N[/tex]. But epsilon is positive and so is x and so is N yet I will get a negative answer for N according to the last statement. Can anyone clarify the argument needed to get [tex] |\frac{-1}{x+1}| = |\frac{1}{x+1}|[/tex] so that N can remain positive? Thanks
From the definition [tex]|f(x)-L|<\epsilon \mbox{ if } x > N, \epsilon>0, N>0[/tex]. It follows that [tex] |\frac{x}{x+1}-1| < \epsilon \mbox{ if } x > N \mbox{ therefore } |\frac{-1}{x+1}| < \epsilon \mbox{ if } x > N[/tex] Now if I move the minus sign across to epsilon I get [tex] |\frac{1}{x+1}| > -\epsilon \mbox{ if } x > N[/tex]. But epsilon is positive and so is x and so is N yet I will get a negative answer for N according to the last statement. Can anyone clarify the argument needed to get [tex] |\frac{-1}{x+1}| = |\frac{1}{x+1}|[/tex] so that N can remain positive? Thanks