Question

Find the limit L for the given function​ f, the point x 0​, and the positive number epsilon. Then find a number delta ​> 0 such​ that, for all​ x, 0less thanStartAbsoluteValue x minus x 0 EndAbsoluteValueless thandelta double right arrow StartAbsoluteValue f (x )minus Upper L EndAbsoluteValueless thanepsilon.

Answer 1

Answer:

L = -25 and  δ = 0.02

Step-by-step explanation:

The function f, point $x_0$ and ε is missing in the question.

The function, f is f(x) = - 4x - 9

           point,  $x_0$ = 4

        epsilon, ε = 0.08

So by the definition of limit,

      $\lim_{x \rightarrow x_0} f(x)= L$

Therefore,

$\lim_{x \rightarrow 4} (-4x-9)$

L= -4(4)-9

L= -16-9

L= -25

So, for every ε > 0, for all δ > 0 such that

|f(x) - L| < ε   $(0<|x-x_0|< \delta)$

$|f(x)-L|<\epsilon \\|(-4x-9)-(-25)|<0.08\\|-4x+16|<0.08\\|-4(x-4)|<0.08\\|-4||x-4|<0.08\\4|x-4|<0.08\\|x-4|<\frac{0.08}{4}\\|x-4|<0.02\\0<|x-4|<0.02 \ \ \ \ \text{ comparing with}\ 0<|x-x_0|< \delta \\ \therefore \delta = 0.02$