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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