Real Analysis: Limits

Limit Laws for Real Functions


In this section, we focus only on limits of real functions using the standard Euclidean metric. This lets us prove a number of properties of limits under very general conditions. These properties are often called the limit laws, and they are very useful because they let us calculate limits of many important real functions at many points.

Algebraic Limit Theorem on $\mathbb{R}$

The algebraic limit theorem lets us quickly solve for the limit of algebraic combinations of functions given a few conservative requirements. If $f$ and $g$ are real functions that both have limits at $c$, then the following properties hold:

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

  • $\lim\limits_{x \rightarrow c}(f(x) + g(x)) = \lim\limits_{x \rightarrow c} f(x) + \lim\limits_{x \rightarrow c} g(x)$ if $c$ is a limit point of $f + g.$

  • $\lim\limits_{x \rightarrow c}(f(x) - g(x)) = \lim\limits_{x \rightarrow c} f(x) - \lim\limits_{x \rightarrow c} g(x)$ if $c$ is a limit point of $f - g.$

  • $\lim\limits_{x \rightarrow c}(f(x)g(x)) = \left(\lim\limits_{x \rightarrow c} f(x)\right)\left(\lim\limits_{x \rightarrow c} g(x)\right)$ if $c$ is a limit point of $fg.$

  • $\lim\limits_{x \rightarrow c}(f(x) / g(x)) = \dfrac{\lim\limits_{x \rightarrow c} f(x)}{\lim\limits_{x \rightarrow c} g(x)}$ if $c$ is a limit point of $f / g$ and $\lim\limits_{x \rightarrow c} g(x) \neq 0.$

Limits of Continuous Real Functions

The continuity limit theorem states that if a function $f$ is continuous at a limit point $c$ in its domain, then $\lim\limits_{x \rightarrow c} f(x) = f(c).$ This theorem directly implies the limit laws for the following families of continuous real functions:

  • Constant functions: $\lim\limits_{x \rightarrow c} a = a.$

  • Monomials: If $m \in \mathcal{M}(\mathbb{R}),$ then $\lim\limits_{x \rightarrow c} m(x) = m(c).$

  • Polynomials: If $p \in \mathcal{P}(\mathbb{R}),$ then $\lim\limits_{x \rightarrow c} p(x) = p(c).$

  • Rational functions: If $p, q \in \mathcal{P}(\mathbb{R})$ and $\lim\limits_{x \rightarrow c} q(x) \neq o,$ then $\lim\limits_{x \rightarrow c} \frac{p(x)}{q(x)} = \frac{p(c)}{q(c)}.$

  • The absolute value function: $\lim\limits_{x \rightarrow c} |x| = |c|.$

One-Sided Limits

Limits require that a function approach a value from both the positive and negative direction around the limit point, assuming the domain is defined there. However, it is possible to define a kind of limit that applies only as the domain values decrease to approach the limit value from the right, or positive direction, or increase to approach it from the left, or negative direction. Such limits are called one-sided limits.

Let $f$ be a real function. The left limit of $f$ as $x$ approaches $c$ equals $L$ if $c$ is a limit point of $\text{Dom}(f) \cap (-\infty, c)$ and for every $\varepsilon > 0$ there exists a $\delta > 0$ such that $|f(x) - L| < \varepsilon$ whenever $0 < c - x < \delta.$ A left limit is written as $\lim\limits_{x \rightarrow c-} f(x) = L.$ Likewise, the right limit of $f$ as $x$ approaches $c$ equals $L$ if If $c$ is a limit point of $\text{Dom}(f) \cap (c, \infty)$, and for every $\varepsilon > 0$ there exists a $\delta > 0$ such that $|f(x) - L| < \varepsilon$ whenever $0 < x - c < \delta.$ A right limit is written as $\lim\limits_{x \rightarrow c+} f(x) = L.$ A one-sided limit is either a left or a right limit. Regular limits are called two-sided limits in order to distinguish them from one-sided limits whenever the context requires it.

Two-Sided Limit Theorem

There is an important and intuitive theorem that relates one-sided limits to two-sided limits. Given a real function $f,$ it states that $\lim\limits_{x \rightarrow c} f(x) = L$ if and only if $\lim\limits_{x \rightarrow c-} f(x) = L = \lim\limits_{x \rightarrow c+}.$ In other words, the two-sided limit exists if and only if both the left and right limits exist and are equal to one another. The stipulation for the equality of the two one-sided limits is key, as the left and the right limits can both exist at $c,$ but if they are not equal, then the two-sided limit will not exist there.

Squeeze Theorem

The squeeze theorem states that if a function is bounded above and below by two other functions in a particular interval, and those functions both converge to the same limit at a point in that interval, then the bounded function also converges to the same limit value at that point. Formally, if $f(x) \leq g(x) \leq h(x)$ on some interval $(a, b),$ and $\lim\limits_{x \rightarrow c} f(x) = \lim\limits_{x \rightarrow c} h(x) = L$ for some $c \in (a, b),$ then $\lim\limits_{x \rightarrow c} g(x) = L.$ The squeeze theorem is useful for when calculating the limit for the bounded function $g$ is difficult, but the calculating the limits for bounding functions $f$ and $g$ are easy to evaluate.


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

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  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.

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

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

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  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) $

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

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

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  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 $\lim\limits_{x \rightarrow c} f(x) = L.$

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

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