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.










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. $\delta = 3ε$.








Explorations

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

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

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

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.
Icons courtesy of icons8.com
For the given function, $f(x) = 2x + 1$, limit, L = 3 and domain value, c = 2, find a value of δ such that $0 < x  c < \delta$ when $f(x)  L < \epsilon$ for the given value of ε = 0.1.
Start by substituting 2x + 1 for f(x), 3 for L and 0.1 for ε in $f(x)  L < \epsilon$ then manipulate the equation until it looks like $0 < x  c < \delta$.
For the given function, $f(x) = 2x + 1$, limit, L = 3 and domain value, c = 2, find a value of δ such that $0 < x  c < \delta$ when $f(x)  L < \epsilon$ for the given value of ε = 0.1.
Start by substituting into the $0 < f(x)  L < \epsilon$ inequality.
$$0 < f(x)  L < \epsilon$$ $$0 < 2x + 1  (3) < 0.1$$ $$0 < 2x + 4 < 0.1$$ $$0 < 2(x  2) < 0.1$$ $$0 < 2(x  2) < 0.1$$ $$0 < 2x  2 < 0.1$$ $$0 < x  2 < 0.05$$So any x value within δ = 0.05 of 2 will have a corresponding function value that's with ε = 0.1 of 3.
For the given function, $f(x) = \sqrt{x}$, limit, L = 3 and domain value, c = 9, find a value of δ such that $0 < x  c < \delta$ when $f(x)  L < \epsilon$ for the given value of ε = 0.2.
Start by substituting 2x + 1 for f(x), 3 for L and 0.1 for ε in $f(x)  L < \epsilon$ then manipulate the equation until it looks like $0 < x  c < \delta$.
For the given function, $f(x) = \sqrt{x}$, limit, L = 3 and domain value, c = 9, find a value of δ such that $0 < x  c < \delta$ when $f(x)  L < \epsilon$ for the given value of ε = 0.2.
Start by substituting into the $0 < f(x)  L < \epsilon$ inequality.
$$0 < f(x)  L < \epsilon$$ $$0 < \sqrt{x}  3 < 0.2$$ $$0.2 < \sqrt{x}  3 < 0.2$$ $$2.8 < \sqrt{x} < 3.2$$ $$2.8^2 < x < 3.2^2$$ $$7.84 < x < 10.24$$So any x value between 7.84 and 10.24 will give us function values with 0.2 of the limit value. However, we need to come up with a δ value that gives us an interval around $x = 9$. $$1.16 < x  9 < 1.24$$
This tells us that the lower bound is 1.16 units from 9 and the upper bound is 1.24 units from 9. Our delta value will be the smaller distance, $\delta = 1.16$.
For the given function, $f(x) = e^x$, limit, L = 1 and domain value, c = 0, find a value of δ such that $0 < x  c < \delta$ when $f(x)  L < \epsilon$ for the given value of ε = 0.1.
Start by substituting into the $0 < f(x)  L < \epsilon$ inequality.
$$0 < f(x)  L < \epsilon$$ $$0 < e^x  1 < 0.1$$ $$0.1 < e^x  1 < 0.1$$ $$0.9 < e^x < 1.1$$ $$ln(0.9) < e^x < ln(1.1)$$ $$0.1053 < x < 0.0953$$The lower bound is 0.1053 units from 0 and the upper bound is 0.0953 units from 0. Our delta value will be the smaller distance, $\delta = 0.0953$.
Show that $\lim\limits_{x\to 3}(3x  1) = 8$ exists by finding a value of δ that corresponds to any value of ε
Start by substituting into the $0 < f(x)  L < \epsilon$ inequality.
$$0 < 3x  1  8 < \epsilon$$ $$0 < 3x  9 < \epsilon$$ $$0 < 3(x  3) < \epsilon$$ $$0 < 3x  3 < \epsilon$$ $$0 < x  3 < \frac{\epsilon}{3}$$So if you chose $\delta = \frac{\epsilon}{3}$ all values of x within that distance from 3 will have function values within ε units of 8.
Show that $\lim\limits_{x\to 0}\sqrt{x + 4} = 2$ exists by finding a value of δ that corresponds to any value of ε
Start by substituting into the $0 < f(x)  L < \epsilon$ inequality.
$$0 < \sqrt{x+4}  2 < \epsilon$$ $$\epsilon < \sqrt{x+4}  2 < \epsilon$$ $$\epsilon + 2 < \sqrt{x+4} < \epsilon + 2$$ $$(\epsilon + 2)^2 < x+4 < (\epsilon + 2)^2$$ $$\epsilon^ 2  4\epsilon + 2^2 < x+4 < \epsilon^ 2 + 4\epsilon + 2^2$$ $$\epsilon^ 2  4\epsilon + 4 < x+4 < \epsilon^ 2 + 4\epsilon + 4$$ $$\epsilon^ 2  4\epsilon < x < \epsilon^ 2 + 4\epsilon$$Since we're looking at the limit as $x o 0 $, the distance from x to 0 must be the minimum of $\epsilon^ 2 + 4\epsilon$ and $\epsilon^ 2 + 4\epsilon.$ Clearly this is $\epsilon^ 2 + 4\epsilon$ so if x is within $\delta = \epsilon^ 2 + 4\epsilon$ of 0 then $f(x)$ will be within ε of 2.
Show that $\lim\limits_{x\to 1}\frac{1}{x} = 1$ exists by finding a value of δ that corresponds to any value of ε.
This one is long so don't be surprised when it doesn't end right away. There are a few things you should keep in mind:
 Remember that if $x < y$ then $\frac{1}{x} > \frac{1}{y}$ or $\frac{1}{y} < \frac{1}{x}$
 To determine if one expressions is greater than or less than another, try setting them equal to each other and solving that equation. If you only get one solution then one of the expressions will always be greater than the other on each side of that value.
 Remember that ε has to be greater than 0.