## 13 Calculus Tutorial Sites, 22 000+ Solved Problems

If you want to learn Calculus, the websites below will most likely help to you. Most of these websites contain conceptual and intuitive explanations of Calculus concepts and most of them are interactive.

All in all, they contain more than 22,000 calculus tutorials and computations.  » Read more

## Limit by epsilon-delta proof: Example 2

This is the overdelayed continuation of the discussion on the $\epsilon-\delta$ definition of limits. In this post, we discuss another example.

Prove that the $\lim_{x \to 2} x^2 = 4$.

Recall that the definition states that the limit of $f(x) = L$ as $x$ approaches $a$ if for all $\epsilon > 0$, however small, there exists a $\delta > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \epsilon$.

From the example 1, we have learned that we should manipulate $|f(x)-L=|x^2 - 4|$, to make one of the expressions look like $|x-a|=|x-2|$. Solving,  we have

$|f(x) - L| = |x^2 - 4| = |(x+2)(x-2)| = |x+2||x-2|$.

Note that we have accomplished our goal, going back to the definition, this means that if $0 < x - 2 < \delta$, then $|x+2||x-2| < \epsilon$.

Now, it is not possible to divide both sides by $x + 2$ (making it $|x-2| < \frac{\epsilon}{|x+2|})$ because $x$ varies. This means that we have to find a constant $k$ such that $|x + 2| < k$. » Read more

## Limit by epsilon-delta proof: Example 1

We have discussed extensively the meaning of the $\epsilon-\delta$ definition.  In this post, we are going to learn some strategies to prove limits of functions by definition.  The meat of the proof is finding a suitable $\delta$ for all possible $\epsilon$ values.

Recall that the definition states that the limit of $f(x) = L$ as $x$ approaches $a$, if for all $\epsilon > 0$, however small, there exists a $\delta > 0$ such that if $0 < | x - a| < \delta$, then $|f(x) - L| < \epsilon$.

Example 1: Let $f(x) = 3x + 5$.  Prove that $\lim_{x \to 2} f(x) = 11$

If we are going to study definition limit above, and apply it to the given function, we have $\lim_{x \to 2} 3x + 5 = 11$, if for all $\epsilon > 0$, however small, there exists a $\delta > 0$ such that if  $0 < |x - 2| < \delta$, then $|3x + 5 - 11| < \epsilon$.  We want to find the value of $\delta$, in terms of $\epsilon$; therefore, we can manipulate one of the inequalities to the other’s form.  In particular, we will manipulate $|3x + 5 - 11| < \epsilon$ to an expression such that the expression inside the absolute value sign will become $x - 2$.

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