Real Analysis: Limits

Problems

1. Let $f : \mathbb{R} \rightarrow \mathbb{R}$ where $f(x) = x.$ Show that $\lim\limits_{x \rightarrow c} f(x) = c.$

   First, note that $c$ is a limit point of $\mathbb{R}.$ Pick $\varepsilon > 0.$ We would like to find a $\delta > 0$ such that $|f(x) - c| < \varepsilon$ whenever $0 < |x - c| < \delta.$ Pick $\delta = \varepsilon.$ Then whenever $0 < |x - c| < \delta,$ it follows that $|f(x) - c| = |x - c| < \delta = \varepsilon.$ Therefore $\lim\limits_{x \rightarrow c} f(x) = c.$

   Show Answer

2. The algebraic limit theorem requires that the limit points of the original functions $f$ and $g$ be a limit point of the algebraic combination. Why is this a necessary stipulation? Give an example of where the limit is defined at $f$ and $g$ but not for some algebraic combination of them.

   The domain of the sum of two functions is the intersection of their domains. For example, if $f(x) = \sqrt{-x}$ and $g(x) = \sqrt{x},$ the domain of $f$ is $(-\infty, 0]$ and the domain of $g$ is $[0, \infty).$ However, domain of $f + g$ is $(-\infty, 0] \cap [0, \infty) = \{0\}.$ Since $f + g$ has finite domain, $0$ is an isolated point and therefore not a limit point.

   Show Answer

3. Algebraic Limit Theorem I: Show that $\lim\limits_{x \rightarrow c} a f(x) = a\lim\limits_{x \rightarrow c} f(x).$

   Assume $\lim\limits_{x \rightarrow c} f(x) = L.$ We would like to show that $\lim\limits_{x \rightarrow c} a f(x) = a\lim\limits_{x \rightarrow c} f(x).$ First, note that if $a = 0,$ then

   $ \lim\limits_{x \rightarrow c} af(x) = \lim\limits_{x \rightarrow c} 0 \\ \lim\limits_{x \rightarrow c} af(x) = 0 \\ \lim\limits_{x \rightarrow c} af(x) = 0 \cdot \lim\limits_{x \rightarrow c} f(x) \\ \lim\limits_{x \rightarrow c} af(x) = a \cdot \lim\limits_{x \rightarrow c} f(x)$

   Otherwise, assume $a \neq 0,$ and expand the metric:

   $ d(af(x), aL) = |af(x) - aL| \\ d(af(x), aL) = |a||f(x) - L| \\ d(af(x), aL) = ad(f(x), L)$

   Pick $\frac{1}{a}\varepsilon > 0.$ Then there is some $\delta > 0$ such that $d(f(x), L) < \frac{1}{a}\varepsilon$ whenever $d(x, c) < \delta.$ Incorporating this into the equation gives the desired reuslt:

   $d(af(x), aL) < a\left(\dfrac{1}{a}\varepsilon\right) \\ d(af(x), aL) < \varepsilon $

   Therefore $\lim\limits_{x \rightarrow c} af(x) = a\lim\limits_{x \rightarrow c}f(x).$

   Show Answer

4. Algebraic Limit Theorem II: Show that if $f$ and $g$ are real functions, $\lim\limits_{x \rightarrow c} f(x) = a$ and $\lim\limits_{x \rightarrow c} g(x) = b,$ and $c$ is a limit point of $\text{Dom}(f + g),$ then $\lim\limits_{x \rightarrow c} (f(x) + g(x)) = \lim\limits_{x \rightarrow c} f(x) + \lim\limits_{x \rightarrow c} g(x).$

   Let $f$ and $g$ be real functions, $\lim\limits_{x \rightarrow c} f(x) = a$ and $\lim\limits_{x \rightarrow c} g(x) = b,$ and $c$ be a limit point of $\text{Dom}(f + g).$ We would like to show that for every $\varepsilon > 0$ there exists a $\delta > 0$ such that $d(f(x) + g(x), a + b) < \varepsilon$ whenever $d(x, c) < \delta.$ First, expand and reorganize the lefthand side of the $\varepsilon$ inequality:

   $ d(f(x) + g(x), a + b) = |(f(x) + g(x)) - (a + b)| \\ d(f(x) + g(x), a + b) = |(f(x) - a) + (g(x) - b)|$

   Now apply the following absolute value inequality:

   $d(f(x) + g(x), a + b) \leq |(f(x) - a)| + |(g(x) - b)|$

   Since $\lim\limits_{x \rightarrow c} f(x) = a,$ it follows that for every $\varepsilon_f > 0$ there exists a $\delta_f > 0$ such that $d(f(x), a) = |f(x) - a| < \varepsilon_f$ whenever $d(x, c) = |x - c| < \delta_f.$ Likewise, since $\lim\limits_{x \rightarrow c} g(x) = b,$ it follows that for every $\varepsilon_g > 0$ there exists a $\delta_g > 0$ such that $d(g(x), b) = |g(x) - b| < \varepsilon_g$ whenever $d(x, c) = |x - c| < \delta_g.$ Pick $\varepsilon_f = \frac{1}{2}\varepsilon$ and $\varepsilon_g = \frac{1}{2}\varepsilon.$ Select $\delta = \text{min}(\{\delta_f, \delta_g\}).$ Then we can make the following substitution:

   $ d(f(x) + g(x), a + b) < \dfrac{1}{2}\varepsilon + \dfrac{1}{2}\varepsilon \\ d(f(x) + g(x), a + b) < \varepsilon $

   Therefore $\lim\limits_{x \rightarrow c}(f(x) + g(x)) = \lim\limits_{x \rightarrow c}f(x) + \lim\limits_{x \rightarrow c}g(x).$

   Show Answer

5. Algebraic Limit Theorem III: Show that if $f$ and $g$ are real functions and $c$ is a limit point of $\text{Dom}(f - g),$ then $\lim\limits_{x \rightarrow c}(f(x) - g(x)) = \lim\limits_{x \rightarrow c}f(x) - \lim\limits_{x \rightarrow c}g(x).$

   Let $f$ and $g$ be real functions, $\lim\limits_{x \rightarrow c}f(x) = a$ and $\lim\limits_{x \rightarrow c}g(x) = b,$ and $c$ be a limit point of $\text{Dom}(f - g).$ Then

   $ \lim\limits_{x \rightarrow c} \left(f(x) - g(x)\right) = \lim\limits_{x \rightarrow c} \left(f(x) + (-g(x))\right) \\ \lim\limits_{x \rightarrow c} f(x) - g(x) = \lim\limits_{x \rightarrow c} f(x) + \lim\limits_{x \rightarrow c}(-g(x)) \\ \lim\limits_{x \rightarrow c} f(x) - g(x) = \lim\limits_{x \rightarrow c} f(x) - \lim\limits_{x \rightarrow c}g(x) $

   Show Answer

6. Algebraic Limit Theorem IV: Show that if $f$ and $g$ are real functions with limits at $c$ and $c$ is a limit point of $\text{Dom}(fg),$ then $\lim\limits_{x \rightarrow c}f(x)g(x) = \left(\lim\limits_{x \rightarrow c}f(x)\right)\left(\lim\limits_{x \rightarrow c}g(x)\right).$

   Assume that $\lim\limits_{x \rightarrow c}f(x) = a$ and $\lim\limits_{x \rightarrow c}g(x) = b,$ and also assume that $c$ is a limit point of $\text{Dom}(fg).$

   We would like to show that for every $\varepsilon > 0,$ there exists a $\delta > 0$ such that $d(f(x)g(x), ab) < \varepsilon$ whenever $0 < d(x, c) < \delta.$ First, take the product of the terms we evaluate for each individual limit of $f$ and $g:$

   $ (f(x) - a)(g(x) - b) = f(x)g(x) - bf(x) - ag(x) + ab $

   Now solve for $f(x)g(x) - ab,$ the expression we would like to evaluate for the limit of the products:

   $ f(x)g(x) - ab = (f(x) - a)(g(x) - b) + bf(x) + ag(x) - 2ab $

   Rearrange the righthand side to be fully in terms of the differences between the functions and their limit values:

   $ f(x)g(x) - ab = (f(x) - a)(g(x) - b) + bf(x) + ag(x) - 2ab \\ f(x)g(x) - ab = (f(x) - a)(g(x) - b) + bf(x) - ab + ag(x) - ab \\ f(x)g(x) - ab = (f(x) - a)(g(x) - b) + b(f(x) - a) + a(g(x) - b) $

   Take the absolute value of both sides and apply inequalities to isolate the limit terms:

   $ |f(x)g(x) - ab| = |(f(x) - a)(g(x) - b) + b(f(x) - a) + a(g(x) - b)| \\ |f(x)g(x) - ab| \leq |(f(x) - a)(g(x) - b)| + |b(f(x) - a)| + |a(g(x) - b)| \\ |f(x)g(x) - ab| \leq |f(x) - a||g(x) - b| + |b||f(x) - a| + |a||g(x) - b| $

   Because $\lim\limits_{x \rightarrow c}f(x) = a$ and $\lim\limits_{x \rightarrow c}g(x) = b,$ if we first assume that $|a| \neq 0$ and $|b| \neq 0,$ we can make the following claims:

   • For every $\sqrt{\frac{\varepsilon}{2}} > 0,$ there exists a $\delta_{f_1} > 0$ such that $|f(x) - a| < \sqrt{\frac{\varepsilon}{2}}$ whenever $0 < |x - c| < \delta_{f_1}.$

   • For every $\sqrt{\frac{\varepsilon}{2}} > 0,$ there exists a $\delta_{g_1} > 0$ such that $|g(x) - b| < \sqrt{\frac{\varepsilon}{2}}$ whenever $0 < |x - c| < \delta_{g_1}.$

   • If $b \neq 0,$ then for every $\frac{\varepsilon}{4|b|} > 0,$ there exists a $\delta_{f_2}$ such that $|f(x) - a| < \frac{\varepsilon}{4|b|}$ whenever $0 < |x - c| < \delta_{f_2},$ and therefore $|b||f(x) - a| < \frac{\varepsilon}{4}.$ If $b = 0,$ then note that $|b||f(x) - a| = 0 < \frac{\varepsilon}{4}.$

   • If $a \neq 0,$ then for every $\frac{\varepsilon}{4|a|} > 0,$ there exists a $\delta_{g_2}$ such that $|g(x) - b| < \frac{\varepsilon}{4|a|}$ whenever $0 < |x - c| < \delta_{g_2},$ and therefore $|a||g(x)-b| < \frac{\varepsilon}{4}.$ If $a = 0,$ then note that $|a||g(x) - b| = 0 < \frac{\varepsilon}{4}.$

   Take $\delta = \text{min}\{ \delta_{f_1}, \delta_{g_1}, \delta_{f_2}, \delta_{g_2} \}.$ Then we can make the follow substitution for when $0 < |x - c| < \delta:$

   $ |f(x)g(x) - ab| < \left(\sqrt{\dfrac{\varepsilon}{2}}\right)\left(\sqrt{\dfrac{\varepsilon}{2}}\right) + \dfrac{\varepsilon}{4} + \dfrac{\varepsilon}{4} \\ |f(x)g(x) - ab| < \dfrac{\varepsilon}{2} + \dfrac{\varepsilon}{4} + \dfrac{\varepsilon}{4} \\ |f(x)g(x) - ab| < \varepsilon $

   Therefore $\lim\limits_{x \rightarrow c}f(x)g(x) = \left(\lim\limits_{x \rightarrow c}f(x)\right)\left(\lim\limits_{x \rightarrow c}g(x)\right).$

   Show Answer

7. Algebraic Limit Theorem V: Show that if $g$ is a real function and $\lim\limits_{x \rightarrow c}g(x) = L \neq 0,$ then $\lim\limits_{x \rightarrow c} \frac{1}{g(x)} = \frac{1}{\lim\limits_{x \rightarrow c}g(x)}.$

   Assume that $g$ is a real function and $\lim\limits_{x \rightarrow c}g(x) = L \neq 0.$ We must associate the difference of the reciprocals $\frac{1}{g(x)}$ and $\frac{1}{L}$ with the difference of the original values $g(x)$ and $L.$ Consider the difference of the reciprocals:

   $ \dfrac{1}{g(x)} - \dfrac{1}{L} $

   Multiplying the first term by $\frac{L}{L}$ and the second term by $\frac{g(x)}{g(x)}$ brings out the $L - g(x)$ term in the numerator:

   $ \dfrac{1}{g(x)} - \dfrac{1}{L} = \dfrac{L}{Lg(x)} - \dfrac{g(x)}{g(x)L} \\ \dfrac{1}{g(x)} - \dfrac{1}{L} = \dfrac{L - g(x)}{Lg(x)} $

   Taking the absolute value of both sides lets us use the $\varepsilon$ bound for the limit of $g$:

   $ \left| \dfrac{1}{g(x)} - \dfrac{1}{L} \right| = \left| \dfrac{L - g(x)}{Lg(x)} \right| \\ \left| \dfrac{1}{g(x)} - \dfrac{1}{L} \right| = \dfrac{|L - g(x)|}{|Lg(x)|} \\ \left| \dfrac{1}{g(x)} - \dfrac{1}{L} \right| = \dfrac{|g(x) - L|}{|Lg(x)|} \\ \left| \dfrac{1}{g(x)} - \dfrac{1}{L} \right| = \dfrac{|g(x) - L|}{|L||g(x)|} $

   The righthand side contains the term $|g(x) - L|$ in the numerator, which will be less than any $\varepsilon > 0$ when $0 < |x - c| < \delta_1$ for some $\delta_1.$ However, the denominator contains a $g(x)$ term in it. This is a problem, since solving for the limit of the reciprocal is the exact thing we are trying to do. However, if we can bound this $g(x)$ term somwhow, we will be able to replace it with a constant.

   Since $\lim\limits_{x \rightarrow c}g(x) = L,$ for every $\varepsilon > 0$ there exists a $\delta_1 > 0$ such that $|g(x) - L| < \varepsilon$ whenever $0 < |x - c| < \delta_1.$ Pick $\varepsilon = \frac{1}{2}L.$ It follows that $|g(x) - L| < \frac{1}{2}|L|.$ However, rather than substitute this inequality in for the numerator, note that it implies that $|g(x)| > \frac{1}{2}L.$ Critically, observe that this also ensures that $g(x) \neq 0$ whenever $0 < |x - c| < \delta_1.$ We substitute this value into the denominator instead:

   $ \left| \dfrac{1}{g(x)} - \dfrac{1}{L} \right| < \dfrac{2|g(x) - L|}{|L||L|} \\ \left| \dfrac{1}{g(x)} - \dfrac{1}{L} \right| < \dfrac{2|g(x) - L|}{|L|^2} $

   Now pick $\frac{1}{2}|L|^2\varepsilon > 0.$ It follows that there is some $\delta_2 > 0$ such that $|g(x) - L| < \frac{1}{2}L^2\varepsilon$ whenever $0 < |x - c| < \delta_2.$ Take $\delta = \text{min}\{\delta_1, \delta_2\}.$ It follows that whenever $0 < |x - c| < \delta,$ the inequality simplifies:

   $ \left| \dfrac{1}{g(x)} - \dfrac{1}{L} \right| < \dfrac{2|L|^2\varepsilon}{2|L|^2} \\ \left| \dfrac{1}{g(x)} - \dfrac{1}{L} \right| < \varepsilon $

   Therefore $\lim\limits_{x \rightarrow c} \frac{1}{g(x)} = \frac{1}{L}.$

   Show Answer

8. Two-Sided Limit Theorem: Given a real function $f,$ show that $\lim\limits_{x \rightarrow c} f(x) = L$ if and only if $\lim\limits_{x \rightarrow c-}f(x) = L = \lim\limits_{x \rightarrow c+} f(x).$

   First, assume that $\lim\limits_{x \rightarrow c}f(x) = L.$ Then $c$ is a limit point of $\text{Dom}(f)$ and for all $\varepsilon > 0$ there exists a $\delta > 0$ such that $d(f(x), L) < \varepsilon$ whenever $0 < d(x, c) < \delta.$ Expanding the definition of the metric on shows that $d(f(x), L) < \varepsilon$ whenever $0 < |x - c| < \delta.$ Expanding the absolute value shows that the condition holds when both $0 < c - x < \delta$ and $0 < x - c < \delta.$ However, these are the conditions for the left and right limits, respectively. Therefore $\lim\limits_{x \rightarrow c-} f(x) = L = \lim\limits_{x \rightarrow c+} f(x).$

   Conversely, assume that $\lim\limits_{x \rightarrow c-} f(x) = L = \lim\limits_{x \rightarrow c+} f(x).$ Then $c$ is a limit point of $\text{Dom}(f) \cap (-\infty, c)$ and $\text{Dom}(f) \cap (c, \infty),$ and so $c$ is a limit point of $\text{Dom}(f).$ Furthermore, for every $\varepsilon > 0$ there exists some $\delta_1 > 0$ such that $d(f(x), L) < \varepsilon$ whenever $0 < c - x < \delta_1,$ and there exists some $\delta_2 > 0$ such that $d(f(x), L) < \varepsilon$ whenever $0 < x - c < \delta_2.$ Take $\delta = \text{min}\{\delta_1, \delta_2\}.$ Then $d(f(x), L) < \varepsilon$ whenever $0 < |x - c| < \delta.$ Therefore $\lim\limits_{x \rightarrow c} f(x) = L.$

   Show Answer

9. Squeeze Theorem: Show that if $f(x) \leq g(x) \leq h(x)$ in some interval $(a, b)$ and $\lim\limits_{x \rightarrow c} f(x) = L = \lim\limits_{x \rightarrow c} h(x)$ for some $c \in (a, b),$ then $\lim\limits_{x \rightarrow c} g(x) = L.$

   Assume $f(x) \leq g(x) \leq h(x)$ along some interval $(a, b),$ and assume that $\lim\limits_{x \rightarrow c} f(x) = L = \lim\limits_{x \rightarrow c} h(x)$ for some $c \in (a, b).$ It follows that for every $\varepsilon > 0,$ there exists a $\delta_f > 0$ where $\delta_f < \text{min}\{|a-c|, |b-c|\}$ such that $|f(x) - L| < \varepsilon$ whenever $0 < |x - c| < \delta_f.$ Therefore $-\varepsilon < f(x) - L < \varepsilon,$ and so $L - \varepsilon < f(x) < L + \varepsilon.$ Likewise, for every $\varepsilon  > 0,$ there exists a $\delta_g > 0$ where $\delta_g < \text{min}\{|a-c|, |b-c|\}$ such that $|g(x) - L| < \varepsilon$ whenever $0 < |x - c| < \delta_g.$ Therefore $-\varepsilon < h(x) - L < \varepsilon,$ and so $L - \varepsilon < h(x) < L + \varepsilon.$ Pick $\delta = \text{min}\{\delta_f, \delta_g\}.$ Then $|f(x) - L| < \varepsilon$ and $|g(x) - L| < \varepsilon$ whenever $0 < |x - c| < \delta.$ It follows that $L - \varepsilon < f(x) \leq g(x) \leq h(x) < L + \varepsilon,$ and so $|g(x) - L| < \varepsilon.$ Therefore $\lim\limits_{x \rightarrow c} g(x) = L.$

   Show Answer