Prove the following statement using the epsilon-delta definition of limit: $\lim\limits_{x \to 3^+} \dfrac{1}{x - 3} = \infty$

Neetesh Kumar | March 06, 2025

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Step-by-step solution:

We need to prove that:

$\lim\limits_{x \to 3^+} \dfrac{1}{x - 3} = \infty$

This means that for any large number $M > 0$, there exists a $\delta > 0$ such that for all $x$ satisfying $3 < x < 3 + \delta$, we have:

$\dfrac{1}{x - 3} > M$

Step 1: Given $M > 0$, we need to find a $\delta > 0$.

Start by manipulating the inequality $\dfrac{1}{x - 3} > M$:

$\dfrac{1}{x - 3} > M$

Take the reciprocal of both sides (which reverses the inequality, because both sides are positive):

$x - 3 < \dfrac{1}{M}$

Thus:

$x < 3 + \dfrac{1}{M}$

Step 2: Choose $\delta$ such that $3 < x < 3 + \delta$ implies the above inequality.

Let:

$\delta = \dfrac{1}{M}$

Now, if $0 < x - 3 < \delta$, it follows that:

$x < 3 + \dfrac{1}{M}$

Therefore, if $x$ is within the interval $(3, 3 + \delta)$, we have:

$\dfrac{1}{x - 3} > M$

Step 3: Conclusion

Thus, for every $M > 0$, we can choose $\delta = \frac{1}{M}$ such that for all $x$ in the interval $(3, 3 + \delta)$, we have:

$\dfrac{1}{x - 3} > M$

This proves that:

$\lim\limits_{x \to 3^+} \dfrac{1}{x - 3} = \infty$

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Final Answer:

$\lim_{x \to 3^+} \dfrac{1}{x - 3} = \infty$

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