Explorations>Lectures>Practice

# Exercises

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## Practice Problems

For the given function, f, limit, L and domain value, c, find a value of δ such that $0 < |x - c| < \delta$ when $|f(x) - L| < \epsilon$ for the given value of ε.

 $f(x) = -2x + 1, c = 2, L = -3, \epsilon = 0.1$ $f(x) = 3x + 4, c = -1, L = 1, \epsilon = 0.1$ $f(x) = \sqrt x, c = 9, L = 3, \epsilon = 0.2$ $f(x) = 2\sqrt{x + 1}, c = 15, L = 8, \epsilon = 0.25$ $f(x) = e^x, c = 0, L = 1, \epsilon = 0.1$ $f(x) = 2^{x+1}, c = -1, L = 0, \epsilon = 0.25$ $f(x) = x^2, c = 2, L = 4, \epsilon = 0.2$ $f(x) = x^2, c = 2, L = 4, \epsilon = 0.1$ $f(x) = x^2 - 1, c = 3, L = 8, \epsilon = 0.1$ $f(x) = \frac{1}{x}, c = 1, L = 1, \epsilon = 0.25$

Show that the following limits exist by finding a value of δ that corresponds to any value of ε. Your δ values will be a function of ε, e.g. $\delta = 3ε$.

 $\lim\limits_{x\to 3}(3x - 1) = 8$ $\lim\limits_{x\to -2}(-x + 1) = 1$ $\lim\limits_{x\to 0}\sqrt{x + 4} = 2$ $\lim\limits_{x\to -2}2\sqrt{2 - x} = 4$ $\lim\limits_{x\to -1}\frac{1}{x} = -1$ $\lim\limits_{x\to 1}-\frac{1}{x+1} = -\frac{1}{2}$ $\lim\limits_{x\to 3}\frac{x^2 - 4}{x+2} = 1$ $\lim\limits_{x\to 0}\frac{x^2 - 1}{x-1} = 1$

## Explorations

1. If $\lim\limits_{x\to a}f(x) = L$, show that $\lim\limits_{x\to a}(f(x) + b) = L + b$ using the ε-δ definition, i.e. find a formula for δ in terms of ε.

2. If $\lim\limits_{x\to c}f(x) = a$, show that $\lim\limits_{x\to 0}f(x+c) = a$.

3. If $\lim\limits_{x\to 0}f(x+c) = a$, show that $\lim\limits_{x\to c}f(x) = a$.

4. The graph to the right is the graph of the function $y = f(x)$. Explain why $\lim\limits_{x\to 1} f(x)$ doesn't exist.

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