finding negative number N so that f(x) - L <e

**zalar81957** · August 1st 2012, 06:58 AM

A positive number e and the limit L of a function f at -infinity are given. Find a negative number N such that ABS(f(x) -L) <e if x<N

lim
x approaches - infinty of 1/(x+19) =0
e= .008

N=?

**Prove It** · August 1st 2012, 07:10 AM

Originally Posted by zalar81957

A positive number e and the limit L of a function f at -infinity are given. Find a negative number N such that ABS(f(x) -L) <e if x<N

lim
x approaches - infinty of 1/(x+19) =0
e= .008

N=?

I can barely understand your question. Are you asking to prove $\displaystyle \begin{align*} \lim_{x \to -\infty}\frac{1}{x + 19} = 0 \end{align*}$ ? If so, you need to show that $\displaystyle \begin{align*} x < N \implies \left|f(x) - L \right| < \epsilon \end{align*}$ for some $\displaystyle \begin{align*} N < 0, \epsilon > 0 \end{align*}$ .

$\displaystyle \begin{align*} \left|f(x) - L \right| &< \epsilon \\ \left|\frac{1}{x + 19} - 0 \right| &< \epsilon \\ \left|\frac{1}{x + 19}\right| &< \epsilon \\ \frac{1}{|x + 19|} &< \epsilon \\ |x + 19| &> \frac{1}{\epsilon} \\ |x| + |19| \geq |x + 19| &> \frac{1}{\epsilon} \textrm{ by the Triangle Inequality} \\ |x| + 19 &> \frac{1}{\epsilon} \\ |x| &> \frac{1}{\epsilon} - 19 \\ x < -\left(\frac{1}{\epsilon} - 19\right) \textrm{ or } x &> \frac{1}{\epsilon} - 19 \\ \textrm{So we can choose to have } x &< 19 - \frac{1}{\epsilon} \end{align*}$

So we can choose $\displaystyle \begin{align*} N = 19 - \frac{1}{\epsilon} \end{align*}$ and reversing each step will prove $\displaystyle \begin{align*} x < N \implies \left|f(x) - L\right| < \epsilon \end{align*}$

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