20. Q20
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Similar Question 1
<p>a) Find a number <code class='latex inline'>\delta</code> such that if <code class='latex inline'>|x-2| < \delta</code>, then <code class='latex inline'>|4x -8|< \epsilon</code>, where <code class='latex inline'>\epsilon = 0.1</code></p>
Similar Question 2
<p>a) Find a number <code class='latex inline'>\delta</code> such that if <code class='latex inline'>|x-2| < \delta</code>, then <code class='latex inline'>|4x -8|< \epsilon</code>, where <code class='latex inline'>\epsilon = 0.1</code></p>
Similar Question 3
<p>Prove the statement using the <code class='latex inline'>\delta, \epsilon</code>` definition of a limit.</p><p><code class='latex inline'>\displaystyle \lim_{x\to 10} (3- \frac{4}{5}x) = -5 </code></p>
Similar Questions
Learning Path
L1 Quick Intro to Factoring Trinomial with Leading a
L2 Introduction to Factoring ax^2+bx+c
L3 Factoring ax^2+bx+c, ex1
Now You Try
<p>(a) Use numerical and graphical evidence to guess the value of the limit</p><p><code class='latex inline'>\displaystyle \lim _{x \rightarrow 1} \frac{x^{3}-1}{\sqrt{x}-1} </code></p><p>(b) How close to 1 does <code class='latex inline'>\displaystyle x </code> have to be to ensure that the function in part (a) is within a distance <code class='latex inline'>\displaystyle 0.5 </code> of its limit?</p>
<p>Prove the statement using the <code class='latex inline'>\delta, \epsilon</code>` definition of a limit.</p><p><code class='latex inline'>\displaystyle \lim_{x\to -3}(1 - 4x) =13 </code></p>
<p>Prove the statement using the <code class='latex inline'>\delta, \epsilon</code>` definition of a limit.</p><p><code class='latex inline'>\displaystyle \lim_{x\to 3}(1 + \frac{1}{3}x) =2 </code></p>
<p>Prove the statement using the <code class='latex inline'>\delta, \epsilon</code>` definition of a limit.</p><p><code class='latex inline'>\displaystyle \lim_{x\to 2} (x^2 -4x + 5) =1 </code></p>
<p>Prove the statement using the <code class='latex inline'>\delta, \epsilon</code>` definition of a limit.</p><p><code class='latex inline'>\displaystyle \lim_{x\to 10} (3- \frac{4}{5}x) = -5 </code></p>
<p>a) Find a number <code class='latex inline'>\delta</code> such that if <code class='latex inline'>|x-2| < \delta</code>, then <code class='latex inline'>|4x -8|< \epsilon</code>, where <code class='latex inline'>\epsilon = 0.1</code></p>
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