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Find the limit as x goes to 0 of (x + e2x)1/x

Evaluate the limit.

    \[ \lim_{x \to 0} \left( x + e^{2x} \right)^{\frac{1}{x}}. \]


From the definition of the exponential we have

    \[ \left( x + e^{2x} \right)^{\frac{1}{x}} = e^{ \frac{1}{x} \log \left( x + e^{2x} \right)}. \]

So, first we use the expansion of e^x as x \to 0 (page 287 of Apostol) to write

    \[ e^{2x} = 1 + 2x + o(x) \qquad \text{as} \quad x \to 0. \]

Therefore, as x \to 0 we have

    \[ \log (x + e^{2x}) = \log (x + 1  + 2x + o(x)) = \log (1 + 3x + o(x)). \]

Now, since 3x + o(x) \to 0 as x \to 0 we can use the expansion (again, page 287) of \log (1+x) as x \to 0 to write

    \[ \log (1 + 3x + o(x)) = 3x+o(x) + o(3x+o(x)) = 3x + o(x) \qquad \text{as} \quad x \to 0. \]

Therefore, as x \to 0 we have

    \[ \frac{1}{x} \log (x+e^{2x}) = \frac{1}{x}(3x + o(x)) = 3 + \frac{o(x)}{x}. \]

So, getting back to the expression we started with,

    \[ \lim_{x \to 0} \left( x + e^{2x} \right)^{\frac{1}{x}} = \lim_{x \to 0} e^{3 + \frac{o(x)}{x}} = e^3 \lim_{x \to 0} e^{\frac{o(x)}{x}}. \]

But, as in the previous exercise (Section 7.11, Exercise #23) we know \lim_{x \to 0} e^{\frac{o(x)}{x}} = 1. Hence,

    \[ \lim_{x \to 0} \left( x + e^{2x} \right)^{\frac{1}{x}} = e^3. \]

Find the limit as x goes to 1 of x1 / (1-x)

Evaluate the limit.

    \[ \lim_{x \to 1} x^{\frac{1}{1-x}}. \]


First, we write

    \[ x^{\frac{1}{1-x}} = e^{\frac{1}{1-x} \cdot \log x}. \]

From this exercise (Section 7.11, Exercise #4) we know that as x \to we have

    \[ \log x = (x-1) + o((x-1)) \]

Therefore, as x \to 1,

    \[ \frac{\log x}{1-x} = -1 + \frac{o(x-1)}{x-1}. \]

So, we then have

    \begin{align*}  \lim_{x \to 1} x^{\frac{1}{1-x}} &= \lim_{x \to 1}e^{\frac{\log x}{1-x}} \\[9pt]  &= \lim_{x \to 1} e^{-1 + \frac{o(x-1)}{x-1}} \\[9pt]   &= \lim_{x \to 1} e^{-1} \cdot e^{\frac{o(x-1)}{x-1}} \\[9pt]  &= \frac{1}{e} \lim_{x \to 1} e^{\frac{o(x-1)}{x-1}}. \end{align*}

(Here we could say that since the exponential is a continuous function we can bring the limit inside and so this becomes e^0 = 1. I’m not sure we know we can pass limits through continuous functions like that, so we continue on with expanding the exponential as in previous exercises.)
Since \frac{o(x-1)}{x-1} \to 0 as x \to 1 we take the expansion of e^x as x \to 0,

    \begin{align*}  \lim_{x \to 1} e^{\frac{o(x-1)}{x-1}} &= \lim_{x \to 1} \left( 1 + \frac{o(x-1)}{x-1} + o \left( \frac{o(x-1)}{x-1} \right)\right) \\[9pt]  &= 1. \end{align*}

Therefore,

    \[ \lim_{x \to 1} x^{\frac{1}{1-x}} = \frac{1}{e} \lim_{x \to 1} e^{\frac{o(x-1)}{x-1}} = \frac{1}{e}. \]

Find the limit as x goes to 0 of (ax – asin x) / x3

Evaluate the limit.

    \[ \lim_{x \to 0} \frac{a^x - a^{\sin x}}{x^3}. \]


First, we want to get expansions for a^x and a^{\sin x} as x \to 0. For a^x we write a^x = e^{x \log a} and use the expansion (page 287 of Apostol) of e^x. This gives us

    \[ a^x = e^{x \log a} = 1 + (x \log a) + \frac{(\log a)^2}{2} x^2 + \frac{(\log a)^3}{6} x^3 + o(x^3). \]

Next, for a^{\sin x}, again we write a^{\sin x} = e^{\sin x \log a} and then use the expansion for e^x we have

    \[ a^{\sin x} = e^{\sin x \log a} = 1 + (\log a \sin x) + \frac{(\log a)^2}{2} (\sin x)^2 + \frac{(\log a)^3}{6} (\sin x)^3 + o((\sin x)^3). \]

Now, we need use the expansion for \sin x (again, page 287 of Apostol)

    \[ \sin x = x - \frac{x^3}{6} + o(x^4) \]

and substitute this into our expansion of a^{\sin x},

    \begin{align*}  a^{\sin x} &= 1 + (\log a \sin x) + \frac{(\log a)^2}{2} (\sin x)^2 + \frac{(\log a)^3}{6} (\sin x)^3 + o((\sin x)^3) \\[9pt]  &= 1 + (\log a) \left( x - \frac{x^3}{6} + o(x^4) \right) + \frac{(\log a)^2}{2} \left( x - \frac{x^3}{6} + o(x^4) \right)^2 \\  & \qquad + \frac{(\log a)^3}{6} \left( x - \frac{x^3}{6} + o(x^4) \right)^3 + o\left( \left( x - \frac{x^3}{6} + o(x^4) \right)^3 \right) \\[9pt]  &= 1 + (\log a) x + \frac{(\log a)^2}{2} x^2 + \left( -\frac{\log a}{6} + \frac{(\log a)^3}{6} \right) x^3 + o(x^3). \end{align*}

(Again, this is the really nice part of little o-notation. We had lots of terms in powers of x greater than 3, but they all get absorbed into o(x^3), so we don’t actually have to multiply out and figure out what they all were. We only need to figure out the terms for the powers of x up to 3. Of course, the 3 could be any number depending on the situation; we chose 3 in this case because we know that’s what we will want in the limit we are trying to evaluate.)

So, now we have expansions for a^x and a^{\sin x} (in which most of the terms cancel when we subtract) and we can evaluate the limit.

    \begin{align*}  \lim_{x \to 0} \frac{a^x - a^{\sin x}}{x^3} &= \lim_{x \to 0} \frac{\frac{\log a}{6} x^3 + o(x^3)}{x^3} \\[9pt]  &= \lim_{x \to 0} \left(\frac{\log a}{6} + \frac{o(x^3)}{x^3}\right)\\[9pt]  &= \frac{\log a}{6}. \end{align*}

Find all x satisfying equations given in terms of sinh

Let c be the number such that \sinh c = \frac{3}{4}. Find all x that satisfy the given equations.

  1. \log (e^x + \sqrt{e^{2x} + 1}) = c.
  2. \log (e^x - \sqrt{e^{2x} - 1}) = c.

  1. We are given \sinh c = \frac{3}{4}. From the formula for \sinh this means

        \[ \frac{e^c - e^{-c}}{2} = \frac{3}{4}. \]

    Then, from the given equation we have

        \[ \log (e^x + \sqrt{e^{2x} + 1}) = c \quad \implies \quad e^x + \sqrt{e^{2x} + 1} = e^c. \]

    Thus,

        \[ e^{-c} = \frac{1}{e^x + \sqrt{e^{2x} + 1}}. \]

    So, then we have

        \begin{align*}  \frac{3}{4} = \frac{e^c - e^{-c}}{2} &= \frac{1}{2} \left( e^x + \sqrt{e^{2x} + 1}  - \frac{1}{e^x + \sqrt{e^{2x}+1}} \right) \\[9pt]  &= \frac{1}{2} \left( \frac{(e^x + \sqrt{e^{2x}+1})^2 - 1}{e^x + \sqrt{e^{2x}+1}} \right) \\[9pt]  &= \frac{e^{2x} + 2e^x \sqrt{e^{2x}+1} + e^{2x}+1 - 1}{2(e^x + \sqrt{e^{2x}+1})} \\[9pt]  &= \frac{e^{2x} + e^x \sqrt{e^{2x}+1}}{e^x + \sqrt{e^{2x}+1}} \\[9pt]  &= e^x \end{align*}

    Therefore we have

        \[ e^x = \frac{3}{4} \quad \implies \quad x = \log 3 - \log 4 = \log 3 - 2 \log 2.\]

  2. There can be no x which satisfy the given equation. As in part (a), we use the definition of \sinh x to obtain the equation,

        \[ \sinh c = \frac{3}{4} \quad \implies \quad e^c - e^{-c} = \frac{3}{2}. \]

    Next, we use the equation given in the problem to write,

        \begin{align*}  &&\log (e^x - \sqrt{e^{2x} - 1}) &= c \\[9pt]  \implies && e^x - \sqrt{e^{2x} -1} &= e^c \\[9pt] \implies && \frac{(e^x - \sqrt{e^{2x} - 1})(e^x + \sqrt{e^{2x}-1})}{e^x + \sqrt{e^{2x}-1}} &= e^c \\[9pt]  \implies && \frac{e^{2x} - e^{2x} + 1}{e^x + \sqrt{e^{2x}-1}} &= e^c \\[9pt]  \implies && \frac{1}{e^x + \sqrt{e^{2x}-1}} &= e^c. \end{align*}

    Furthermore, we can obtain an expression for e^{-c} by considering

        \[ e^x - \sqrt{e^{2x} - 1} &= e^c \quad \implies \quad \frac{1}{e^x - \sqrt{e^{2x} - 1}} &= e^{-c}. \]

    Putting these expressions for e^c and e^{-c} into our original equation we have

        \begin{align*}  \frac{3}{2} &= e^c - e^{-c} \\[9pt]  &= \left( \frac{1}{e^x + \sqrt{e^{2x}-1}} \right) - \left( \frac{1}{e^x - \sqrt{e^{2x}-1}} \right) \\[9pt]  &= \frac{e^x - \sqrt{e^{2x}-1} - e^x - \sqrt{e^{2x}-1}}{(e^x + \sqrt{e^{2x}-1})(e^x - \sqrt{e^{2x}-1})} \\[9pt]  &= \frac{ -2 \sqrt{e^{2x}-1}}{e^{2x} - e^{2x} + 1} \\[9pt]  &= -2\sqrt{e^{2x}-1} \end{align*}

    But this implies

        \[ \sqrt{e^{2x}-1} = -\frac{3}{4} \]

    which is impossible. Hence, there can be no real x satisfying this equation.

Derive some properties of the product of ex with a polynomial

Let

    \[ f(x) = e^x p(x) \qquad \text{where} \qquad p(x) = c_0 + c_1 x + c_2x^2. \]

  1. Prove that

        \[ f^{(n)} (0) = c_0 + nc_1 + n(n-1)c_2 \]

    where f^{(n)} denotes the nth derivative of f.

  2. Do part (a) in the case that p(x) is a cubic polynomial.
  3. Find a similar formula and prove it in the case that p(x) is a polynomial of degree m.

For all of these we recall from a previous exercise (Section 5.11, Exercise #4) that by Leibniz’s formula if h(x) = f(x)g(x) then the nth derivative h^{(n)}(x) is given by

    \[ h^{(n)} (x) = \sum_{k=0}^n \binom{n}{k} f^{(k)} (x) g^{(n-k)}(x). \]

So, in the case at hand we have f(x) = e^x p(x) and so

    \[ f^{(n)} (x) = \sum_{k=0}^n \binom{n}{k} p^{(k)}(x) e^x. \]

(Since the (n-k)th derivative of e^x is still e^x for all n and k.)

  1. Proof. From the formula above we have

        \[ f^{(n)}(0) = \sum_{k=0}^n \binom{n}{k} p^{(k)}(0) e^0 = \sum_{k=0}^n \binom{n}{k} p^{(k)}(0). \]

    But, since p(x) is a quadratic polynomial we have

        \begin{align*}  p(x) &= c_0 + c_1 x + c_2 x^2 \\[9pt]  p'(x) &= c_1 + 2c_2 x \\[9pt]  p''(x) &= 2c_2 \\[9pt]  p^{(k)} (x) &= 0 & \text{for all } k \geq 3. \end{align*}

    Hence, we have

        \begin{align*}   f^{(n)}(0) &= \binom{n}{0} p(0) + \binom{n}{1} p'(0) + \binom{n}{2} p''(0) \\[9pt]  &= c_0 + n c_1 + \frac{n(n-1)}{2} 2c_2 \\[9pt]  &= c_0 + n c_1 + n(n-1) c_2. \qquad \blacksquare \end{align*}

  2. If p(x) is a cubic polynomial we may write,

        \[ p(x) = c_0 + c_1 x + c_2 x^2 + c_3 x^3. \]

    Claim: If f(x) = e^x p(x) then

        \[ f^{(n)}(0) = c_0 + nc_1 + n(n-1)c_2 + n(n-1)(n-2)c_3. \]

    Proof. We follow the exact same procedure as part (a) except now we have the derivatives of p(x) given by

        \begin{align*}  p(x) &= c_0 + c_1 x + c_2x^2 + c_3x^3 \\[9pt]  p'(x) &= c_1 + 2c_2 x + 3c_3 x^2 \\[9pt]  p''(x) &= 2c_2 + 6c_3 x \\[9pt]  p^{(3)}(x) &= 6c_3 \\[9pt]  p^{(k)}(x) &= 0 & \text{for all } k \geq 4.  \end{align*}

    Therefore, we now have

        \begin{align*}  f^{(n)}(0) &= \sum_{k=0}^n \binom{n}{k} p^{(k)}(0) e^0 \\[9pt]  &= \sum_{k=0}^n \binom{n}{k} p^{(k)}(0) \\[9pt]  &= \binom{n}{0} p(0) + \binom{n}{1} p'(0) + \binom{n}{2} p''(0) + \binom{n}{3} p'''(0) \\[9pt]  &= c_0 + nc_1 + \frac{n(n-1)}{2} 2c_2 + \frac{n(n-1)(n-2)}{6} 6c_3 \\[9pt]  &= c_0 + nc_1 + n(n-1)c_2 + n(n-1)(n-2)c_3. \qquad \blacksquare \end{align*}

  3. Claim: Let p(x) be a polynomial of degree m,

        \[ p(x) = \sum_{k=0}^m c_k x^k. \]

    Let f(x) = e^x p(x). Then,

        \[ f^{(n)}(0) = \sum_{k=0}^m k! \binom{n}{k} c_k. \]

    Proof. Using Leibniz’s formula again, we have

        \[ f^{(n)}(0) = \sum_{k=0}^m \binom{n}{k} p^{(k)}(0). \]

    But for the degree m polynomial p(x), we know p^{(k)}(0) = k!c_k if 0 \leq k \leq m and p^{(k)} (0) = 0 for all k > m. Hence, we have

        \[ f^{(n)}(0) = \sum_{k=0}^m k! \binom{n}{k} c_k. \qquad \blacksquare \]

Prove some properties of a differentiable function satisfying a given functional equation

Let g be a function differentiable everywhere such that

    \[ g'(0) = 2 \quad \text{and} \quad g(x+y) = e^y g(x) + e^x g(y) \qquad \text{for all } x,y \in \mathbb{R}. \]

  1. Prove that g(2x) = 2e^x g(x) and conjecture and prove a similar formula for g(3x).
  2. Conjecture and prove a formula for g(nx) in terms of g(x) for all positive integers n.
  3. Prove that g(0) = 0 and compute

        \[ \lim_{h \to 0} \frac{g(h)}{h}. \]

  4. Prove that there exists a constant C such that

        \[ g'(x) = g(x) + Ce^x \]

    for all x. Find the value of the constant C.


  1. Proof. We can compute this using the functional equation:

        \[ g(2x) = g(x+x) = e^x g(x) + e^x g(x) = 2e^x g(x). \qquad \blacksquare \]

    Next, we conjecture

        \[ g(3x) = 3e^{2x} g(x). \]

    Proof. Again, we compute using the functional equation, and the above formula for g(2x),

        \begin{align*}  g(3x) &= g(2x + x) \\  &= e^x g(2x) + e^{2x}g(x) \\  &= e^x (2e^x g(x)) + e^{2x} g(x) \\  &= 2e^{2x}g(x) + e^{2x}g(x) \\  &= 3e^{2x}g(x). \qquad \blacksquare \end{align*}

  2. We conjecture

        \[ g(nx) = ne^{(n-1)x} g(x). \]

    Proof. The proof is by induction. We have already established the cases n = 2 and n = 3 (and the n = 1 case is the trivial g(x) = g(x)). Assume then that the formula holds for some integer k \geq 1. Then we have

        \begin{align*}  g((k+1)x) &= g(kx + x) \\  &= e^x g(kx) + e^{kx} g(x) &(\text{functional equation})\\  &= e^x (ke^{(k-1)x} g(x)) + e^{kx}g(x) &(\text{induction hypothesis}) \\  &= ke^{kx} g(x) + e^{kx} g(x) \\  &= (k+1)e^{kx} g(x). \end{align*}

    Thus, if the formula holds for k, it also holds for k + 1. Hence, by induction it holds for all integers k. \qquad \blacksquare

  3. Proof. Using the functional equation

        \[ g(0) = g(0+0) = e^0 g(0) + e^0 g(0) = 2g(0) \quad \implies \quad g(0) = 0. \]

    Then, since the derivative g'(x) exists for all x (by hypothesis) we know it must exist in particular at x = 0. Using the limit definition of derivative, and the facts that g(0) = 0 and g'(0) = 2 we have

        \[ g'(0) = \lim_{h \to 0} \frac{g(0+h) - g(0)}{h} = \lim_{h \to 0} \frac{g(h)}{h} = 2. \qquad \blacksquare\]

  4. Proof. Since the derivative g'(x) must exist for all x (by hypothesis) we know that the limit

        \[ \lim_{h \to 0} \frac{g(x+h) - g(x)}{h} \]

    must exist for all x. Using the functional equation for g we have

        \begin{align*}  \lim_{h \to 0} \frac{g(x+h) - g(x)}{h} &= \lim_{h \to 0} \frac{e^h g(x) + e^x g(h) - g(x)}{h} \\[9pt]  &= \lim_{h \to 0} \frac{g(x)(e^h - 1) + e^x g(h)}{h}. \end{align*}

    But then, from part (c) we know \lim_{h \to 0} \frac{g(h)}{h} = 2 and from this exercise (Section 6.17, Exercise #38) we know

        \[ \lim_{x \to 0} \frac{e^x - 1}{x} = 1. \]

    Therefore, we have

        \begin{align*}  \lim_{h \to 0} \frac{g(x)(e^h - 1) + e^x g(h)}{h} &= \lim_{h \to 0} \left( \frac{e^x g(h)}{h} \right) + \lim_{h \to 0} \left( \frac{g(x)(e^h - 1)}{h} \right) \\  &= e^x \lim_{h \to 0} \left(\frac{g(h)}{h} \right) + g(x) \lim_{h \to 0} \left( \frac{e^h-1}{h} \right) \\  &= 2e^x + g(x). \qquad \blacksquare \end{align*}

Prove that a function satisfying given properties must be ex

Given a function f(x) satisfying the properties:

    \[ f(x+y) = f(x) f(y) \qquad \text{for all } x, y \in \mathbb{R}\]

and

    \[ f(x) = 1 + x g(x) \quad \text{where} \quad \lim_{x \to 0} g(x) = 1. \]

Prove the following:

  1. The derivative f'(x) exists for all x.
  2. We must have f(x) = e^x.

This problem is quite similar to two previous exercises here and here (Section 6.17, Exercises #39 and #40).

  1. Proof. To show that the derivative f'(x) exists for all x we must show that the limit

        \[ \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

    exists for all x. Using the given properties of f(x) we can evaluate this limit

        \begin{align*}  \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} &= \lim_{h \to 0} \frac{f(x)f(h) - f(x)}{h} \\[9pt]  &= \lim_{h \to 0} \frac{f(x)(f(h) - 1)}{h} \\[9pt]  &= \lim_{h \to 0} \frac{f(x)(hg(h))}{h} &(\text{using } f(h) = 1 + hg(h)) \\[9pt]  &= \lim_{h \to 0} f(x)g(h) \\[9pt]  &= f(x) &(\text{using } \lim_{h \to 0} g(h) = 1). \end{align*}

    Therefore, f'(x) = f(x) for all x, so the derivative is defined everywhere. \qquad \blacksquare

  2. Proof. From part (a) we know f(x) = f'(x). By Section 6.17, Exercise #39 (linked above) we know that the only functions f(x) which satisfy this equation are f(x) = 0 for all x or f(x) = Ke^x for some constant K (where c = 1 in the linked exercise). However, since the derivative of f exists everywhere, and differentiability implies continuity, we know f is continuous everywhere. Hence, \lim_{x \to 0} f(x) = f(0). Then,

        \[ \lim_{x \to 0} f(x) = \lim_{x \to 0} (1 + xg(x)) = 1 + 0 = 1 \]

    since \lim_{x \to 0} g(x) = 1, so \lim_{x \to 0} xg(x) = 0. Therefore, we must have f(x) = Ke^x for some constant K. Furthermore, we must have K = 1 since f(0) = Ke^0 = K = 1. Thus, f(x) = e^x. \qquad \blacksquare