# Find a number such that for all x, 0 < < δ ⇒ < ε.???? f(x) = 6x - 9, L = -3, x0 = 1, and ε = 0.01 0.001667 0.01 0.003333 0.000833

Oct 14, 2017

The value of $\delta = 0.001667$

#### Explanation:

The definition of the limit is

${\lim}_{x \to {x}_{0}} f \left(x\right) = L$

$\forall \epsilon > 0 , \exists \delta > 0$ such that

$| f \left(x\right) - L | < \epsilon$

when $| x - {x}_{0} | < \delta$

Here,

$f \left(x\right) = 6 x - 9$

${x}_{0} = 1$

$L = - 3$

$\epsilon = 0.01$

Therefore,

$| 6 x - 9 - \left(- 3\right) | < 0.01$

$| 6 x - 6 | < 0.01$

$- 0.01 < 6 x - 6 < 0.01$

$- 0.01 + 6 < 6 x < 0.01 + 6$

$\frac{5.99}{6} < x < \frac{6.01}{6}$

$0.998333 < x < 1.001666$
Also,

$| x - 1 | < \delta$

$- \delta < x - 1 < \delta$

$1 - \delta < x < 1 + \delta$

Therefore,

$1 - \delta = 0.998333$

$\delta = 1 - 0.998333 = 0.001667$

$1 + \delta = 1.001666$

$\delta = 1.001666 - 1 = 0.001667$

Oct 14, 2017

Also, for this example, given any $\epsilon > 0$, the value $\delta = \frac{\epsilon}{6} > 0$ will be such that $| f \left(x\right) - L | < \epsilon$ for all $x$ satisfying $| x - {x}_{0} | < \delta$.

#### Explanation:

If $| x - {x}_{0} | = | x - 1 | < \delta = \frac{\epsilon}{6}$, then

$| f \left(x\right) - L | = | 6 x - 9 + 3 |$

$= | 6 \left(x - 1\right) | = 6 | x - 1 | < 6 \cdot \frac{\epsilon}{6} = \epsilon$.

This proves that ${\lim}_{x \to 1} \left(6 x - 9\right) = - 3$.