- #1

karush

Gold Member

MHB

- 3,269

- 5

prove the statement using the $\epsilon,\delta$ definition of a limit.

$$\lim_{{x}\to{1}}\frac{2+4x}{3}=2$$

so then

$$x_0=1\quad f(x)=\frac{2+4x}{3}\quad L=2$$

now

$$0<|x-1|<\delta\quad\text

{and}\quad\left|\frac{2+4x}{3}-2\right|

<\epsilon$$

then

$$\left|\frac{2+4x}{3}-\frac{6}{3}\right|=\left|\frac{4x-4}{3}\right|$$

$$=\frac{4}{3}|x-1|=|x-1|<\frac{3}{4}\epsilon$$

finally

$$\left|\frac{2+4x}{3}-2\right|

=\frac{4}{3}|x-1|<\frac{4}{3}\delta

=\frac{4}{3}\left(\frac{3}{4}\epsilon\right)

=\epsilon.$$

$$\lim_{{x}\to{1}}\frac{2+4x}{3}=2$$

so then

$$x_0=1\quad f(x)=\frac{2+4x}{3}\quad L=2$$

now

$$0<|x-1|<\delta\quad\text

{and}\quad\left|\frac{2+4x}{3}-2\right|

<\epsilon$$

then

$$\left|\frac{2+4x}{3}-\frac{6}{3}\right|=\left|\frac{4x-4}{3}\right|$$

$$=\frac{4}{3}|x-1|=|x-1|<\frac{3}{4}\epsilon$$

finally

$$\left|\frac{2+4x}{3}-2\right|

=\frac{4}{3}|x-1|<\frac{4}{3}\delta

=\frac{4}{3}\left(\frac{3}{4}\epsilon\right)

=\epsilon.$$

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