Question

Video Example Prove the following limit. $\lim_{x \to 2} (5x - 3) = 7$ SOLUTION 1. Preliminary analysis of the problem (guessing a value for $\delta$). Let $\epsilon$ be a given positive number. We want to find a number $\delta$ such that if $0 < |x - 2| < \delta$ then $|(5x - 3) - 7| < \epsilon$. But $|(5x - 3) - 7| = |5x - 10| = 5|x - 2|$. Therefore, we want $\delta$ such that if $0 < |x - 2| < \delta$ then $5|x - 2| < \epsilon$ that is, if $0 < |x - 2| < \delta$ then $|x - 2| < \frac{\epsilon}{5}$ This suggests that we should choose $\delta = \frac{\epsilon}{5}$ 2. Proof (showing that $\delta$ works). Given $\epsilon > 0$, choose $\delta = \frac{\epsilon}{5}$. If $0 < |x - 2| < \delta$, then we get the following. $|(5x - 3) - 7| = |5x - 10| = 5|x - 2| < 5\delta = 5\left(\frac{\epsilon}{5}\right) = \epsilon$ Thus, if $0 < |x - 2| < \delta$ then $|(5x - 3) - 7| < \epsilon$. Therefore, by the definition of a limit, we get the following. $\lim_{x \to 2} (5x - 3) = 7$ Need Help? Read It

          Video Example
Prove the following limit.
$\lim_{x \to 2} (5x - 3) = 7$
SOLUTION
1. Preliminary analysis of the problem (guessing a value for $\delta$). Let $\epsilon$ be a given positive number. We want to find a
number $\delta$ such that if $0 < |x - 2| < \delta$ then $|(5x - 3) - 7| < \epsilon$.
But $|(5x - 3) - 7| = |5x - 10| = 5|x - 2|$. Therefore, we want $\delta$ such that
if $0 < |x - 2| < \delta$ then $5|x - 2| < \epsilon$
that is, if $0 < |x - 2| < \delta$ then $|x - 2| < \frac{\epsilon}{5}$
This suggests that we should choose $\delta = \frac{\epsilon}{5}$
2. Proof (showing that $\delta$ works). Given $\epsilon > 0$, choose $\delta = \frac{\epsilon}{5}$. If $0 < |x - 2| < \delta$, then we get the
following.
$|(5x - 3) - 7| = |5x - 10| = 5|x - 2| < 5\delta = 5\left(\frac{\epsilon}{5}\right) = \epsilon$
Thus, if $0 < |x - 2| < \delta$ then $|(5x - 3) - 7| < \epsilon$.
Therefore, by the definition of a limit, we get the following.
$\lim_{x \to 2} (5x - 3) = 7$
Need Help? Read It

Video Example
Prove the following limit.
limx → 2 (5x - 3) = 7
SOLUTION
1. Preliminary analysis of the problem (guessing a value for δ). Let ϵ be a given positive number. We want to find a
number δ such that if 0 < |x - 2| < δ then |(5x - 3) - 7| < ϵ.
But |(5x - 3) - 7| = |5x - 10| = 5|x - 2|. Therefore, we want δ such that
if 0 < |x - 2| < δ then 5|x - 2| < ϵ
that is, if 0 < |x - 2| < δ then |x - 2| < (ϵ)/(5)
This suggests that we should choose δ = (ϵ)/(5)
2. Proof (showing that δ works). Given ϵ > 0, choose δ = (ϵ)/(5). If 0 < |x - 2| < δ, then we get the
following.
|(5x - 3) - 7| = |5x - 10| = 5|x - 2| < 5δ = 5((ϵ)/(5)) = ϵ
Thus, if 0 < |x - 2| < δ then |(5x - 3) - 7| < ϵ.
Therefore, by the definition of a limit, we get the following.
limx → 2 (5x - 3) = 7
Need Help? Read It

Added by Jerry W.

Question

Please give Ace some feedback