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Vanilla application of L’Hospital

A3, 2010

Evaluate the limit:

\[ \lim_{x \rightarrow 1}\left(\frac{n- \displaystyle \sum_{k=1}^n x^{k}}{1-x}\right) \]

Solution We apply the L'Hospital's rule and differentiate both the numerator and the denominator. \begin{array}{rl} \lim_{x\rightarrow 1}&\displaystyle \frac{-n x^{n-1}-(n-1) x^{n-2}-\cdots-x-1}{-1} \\ =&\frac{n(n-1)}{2} \end{array}

Practice Problem

Explain what is wrong with the use of L'Hospital's rule. Find the correct limit. \[\lim_{x \rightarrow 1} \frac{2 x^{3}-3 x+1}{x^{4}-1}=\lim_{x \rightarrow 1} \frac{6 x^{2}-3}{4 x^{3}}=\lim_{x \rightarrow 1} \frac{12 x}{12 x^{2}}=\lim_{x \rightarrow 1} \frac{1}{x}=1\]

L’Hospital again

A4, 2012

Prove the following limit.

\[ \lim_{x \rightarrow \infty} \frac{x^{100} \ln (x)}{e^{x} \arctan \left(\frac{\pi}{3}+\sin x\right)}=0 \]


For some positive constant \(k\) we can ensure that \(\arctan \left(\frac{\pi}{3}+\sin x\right)> k\) for any \(x\). For example, \(k=\arctan(0.0001)\) will work. This is because \(\pi>3.142, \sin (x) \geq-1\) and \(\arctan\) is an increasing function.

Further, \(\ln (x)< x\) for \(x> 0\). So the given ratio must lie between 0 and \(x^{101} / c e^{x} \). If we apply the L'Hospital's rule 102 times, we get the result.

Absolute value in the denominator

A6, 2022

Which of the options are true about the function \(f\) as defined below: \[ f(x) = \frac{1}{|\ln x|} \left( \frac{1}{x} + \cos x \right) \]

  1. As \(x\rightarrow \infty\), the sign of \(f\) changes infinitely many times.
  2. \(\lim_{x\rightarrow \infty} f(x) \) does not exist.
  3. \(\lim_{x\rightarrow 1} f(x) = \infty \).
  4. \(\lim_{x\rightarrow 0^+} f(x) = 1\).

  • TRUE.
  • FALSE.
  • TRUE.
  • FALSE.

  • Tower expression

    A7, 2022

    Let \(f_0(x) = x\),\(f_1(x)=x^x\),\(f_2(x) = x^{x^x}\), etc. For \(x>0\) which of the following options are true?

    1. \(\lim_{x\rightarrow 0^+} f_1(x) = 1\)
    2. \(\lim_{x\rightarrow 0^+} f_2(x) = 1\)
    3. \( \int_0^{1} f_3 dx = 1 \)
    4. \( f_{123}^\prime(1) = 1 \)

  • TRUE.
  • FALSE.
  • FALSE.
  • TRUE.

  • Limit of \(x^{x^{x}}\)

    B1a, 2017

    (a) Evaluate \( \lim_{x \rightarrow 0^{+}}\left(x^{x^{x}}-x^{x}\right) \)


    First consider the limit \[ \begin{align} \lim_{x \rightarrow 0^{+}} x^{x} \\ &=\lim_{x \rightarrow 0^{+}}\left(e^{x\log x}\right) \\ &=\lim_{x \rightarrow 0^{+}}\left(e^{\frac{\log x}{1 / x}}\right) \end{align} \]

    Now consider just the exponent: \[ \begin{align} \lim_{x \rightarrow 0^{+}} \frac{\log x}{1 / x} \\ &=\lim_{x \rightarrow 0} \frac{1 / x}{-1 / x^{2}} \\ &=0 \end{align} \] substituting the value 0 from (2) in equation (1) we get that the limit is 1.

    Now \[ \begin{align} \lim_{x \rightarrow 0^{+}}\left(x^{x^{x}}-x^{x}\right) \\ &=\lim_{x \rightarrow 0^{+}} x^{x^{x}}-\lim_{x \rightarrow 0^{+}} x^{x} \\ &=\lim_{x \rightarrow 0^{+}} x^{\lim_{x \rightarrow 0^{+}} x^{x}}-\lim_{x \rightarrow 0^{+}} x^{x} \\ &=0-1 \\ &=-1 \end{align} \]

    Reference. This problem is based on a general result: A tower of even number of \(x\)s tends to zero and a tower of odd number of \(x\)s tends to one [1].
    [1] The limit of x** as x tends to zero J. Marshall Ash, Mathematics Magazine, 69 (1996), 207-209.

    Simple limits

    A9, 2021

    Let \(f\) and \(g\) be function defined as follows:
    \[ f(x) = \frac{x}{x+\sin x} \] \[ g(x) = \frac{x^4+x^6}{e^x-1-x^2} \] (a) \(\lim_{x\rightarrow 0} f(x) = 1/2\)
    (b) \(\lim_{x\rightarrow \infty} f(x)\) does not exist.
    (c) \(\lim_{x\rightarrow \infty} g(x) \) is finite.
    (d) \(\lim_{x\rightarrow 0} g(x) = 720 \).

    Solution (a) True. \(\sin x \approx x\) as \(x\rightarrow 0\).
    (b) False. The limit is 1.
    (c) True. Numerator is polynomial, while the denominator is exponential.
    (d) False. The limit is zero. (L'Hosptial's rule is not applicable).

    Smallest prime factor function

    A9, 2017

    Let \(f\) be a continuous function from \(\mathbb{R}\) to \(\mathbb{R}\) (where \(\mathbb{R}\) is the set of all real numbers) that satisfies the following property:

    For every natural number \(n\)

    \[ f(n)= \text{the smallest prime factor of }n \]

    For example, \(f(12)=2, f(105)=3 .\) Calculate the following.

    (a) \(\lim_{x \rightarrow \infty} f(x)\)

    (b) The number of solutions to the equation \(f(x)=2016\)


    (a) \(f(x)\) will take value 2 for all even \(x\). At the same time, primes provide an increasing infinite sequence of positive integers for which \(f(x)=x .\) Thus \(\lim_{x \rightarrow \infty} f(x)\) does not exist.

    (b) By intermediate value theorem, for each prime \(p> 2016\) there is an \(x\) between \(p\) and \(p+1\) such that \(f(x)=2016\)

    Limit of a log of an exponent

    A9, 2019

    Consider \(f: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}\) defined as follows:

    \[ f(a, b):=\lim_{n \rightarrow \infty} \frac{1}{n} \log_{e}\left[e^{n a}+e^{n b}\right] \]

    (a) \(f\) is not onto i.e. the range of \(f\) is not all of \(\mathbb{R}\).

    (b) For every \(a\) the function \(x \mapsto f(a, x)\) is continuous everywhere.

    (c) For every \(b\) the function \(x \mapsto f(x, b)\) is differentiable everywhere.

    (d) We have \(f(0, x)=x\) for all \(x \geq 0\)


    Without loss of generality, assume that \(a > b\).

    \begin{align*} f(a, b) &=\lim_{n \rightarrow \infty} \frac{1}{n} \log\left(e^{n a}+e^{n b} \right) \\ & =\lim_{n \rightarrow \infty} \frac{1}{n} \log \left( e^{na}( 1 +e^{n(b-a)} ) \right) \\ & =\lim_{n \rightarrow \infty} \frac{1}{n} \log e^{na} + \log ( 1 +e^{n(b-a)} )\\ & = a + \lim_{n\rightarrow \infty} \frac{1}{n} \log ( 1 + e^{-\infty} ) \\ & = a + \lim_{n\rightarrow \infty} \frac{1}{n} \log ( 1 )\\ & = a \end{align*}

    Similary, if \(b \geq a\), \(f (a,b) = b \). Thus, we have: \[ f(a,b) = \max(a,b) \]

    1. False. \(f(x,x) = x \) so \(f\) is onto.
    2. True. \(g(x) = \max(x,a) \) which is continuous everywhere.
    3. False. \( h(x) = \max(x,b) \) is not differentiable at \(x = b\). More explicitly, \( \lim_{x \rightarrow b^+} = 1 \) and \( \lim_{x \rightarrow b^-} = 0 \).
    4. True. \( \max(x,0) = x \) for all \(x \geq 0 \).

    Polynomial and limits

    B1, 2015

    (a) For any polynomial \(p(t),\) the limit \(\lim_{t \rightarrow \infty} \frac{p(t)}{e^{t}}\) is independent of the polynomial \(p .\) Justify this statement and find the value of the limit.

    (b) Consider the function defined by

    \begin{align} q(x) &=e^{-1 / x} \text { when } x>0 \\ &=0 \text { when } x=0 \\ &=e^{1 / x} \text { when } x<0 \end{align}

    Show that \(q^{\prime}(0)\) exists and find its value. Why is it enough to calculate the relevant limit from only one side?

    (c) Now for any positive integer \(n,\) show that \(q^{(n)}(0)\) exists and find its value. Here \(q(x)\) is the function in part (b) and \(q^{(n)}(0)\) denotes its \(n\) -th derivative at \(x=0\).


    (a) If \(p(t)\) is constant, then the limit \(=0 .\) Otherwise we get a form \(\frac{\pm \infty}{\infty}\). Using L'Hospital's rule, we get \(\lim_{t \rightarrow \infty} \frac{p(t)}{e^{t}}=\lim_{t \rightarrow \infty} \frac{p^{\prime}(t)}{e^{t}}=0\) by induction on the degree of \(t\) (or apply L'Hospital's rule repeatedly).

    (b) The right side derivative \(=\lim_{h \rightarrow 0^{+}} \frac{q(h)-q(0)}{h}=\lim_{h \rightarrow 0^{+}} \frac{e^{-1 / h}}{h}=\lim_{h \rightarrow 0^{+}} \frac{1 / h}{e^{1 / h}}=\lim_{t \rightarrow+\infty} \frac{t}{e^{t}} \cdot\) (Let \(t=1 / h .\) As \(\left.h \rightarrow 0^{+}, t \rightarrow+\infty .\right)\) This limit is \(0,\) e.g. by part \((\mathrm{a})\) Now \(q\) is an even function, so letting \(k=-h,\) the left side derivative \(=\lim_{h \rightarrow 0^{-}} \frac{q(h)-q(0)}{h}=\) \(\lim_{k \rightarrow 0^{+}} \frac{q(-k)}{-k}=\lim_{k \rightarrow 0^{+}} \frac{q(k)}{-k} .\) Using the earlier calculation this also equals \(-0=0\) Note: It is wrong to argue that \(q^{\prime}(0)=\lim_{x \rightarrow 0} q^{\prime}(x)\) because to do so, we first need to know that \(q^{\prime}\) is continuous at \(0,\) but we have not even shown that \(q^{\prime}(0)\) exists! For the same reason it is wrong to argue below that \(q^{(n)}(0)=\lim_{x \rightarrow 0} q^{(n)}(x)\)

    (c) We will show by induction on \(n\) that \(q^{(n)}(0)=0 .\) The case \(n=1\) is done. (We can even start the induction at \(n=0\) by interpreting \(q^{(0)}(x)=q(x) .\) ) Assuming that we are done up to \(n\) and to prove the statement for \(n+1,\) we need to calculate \(\lim_{h \rightarrow 0} \frac{q^{(n)}(h)-q^{(n)}(0)}{h}=\) \(\lim_{h \rightarrow 0} \frac{q^{(n)}(h)}{h},\) because \(q^{(n)}(0)=0\) by induction. Therefore it is good to examine \(q^{(n)}(h)\) for \(h \neq 0 .\) This is easy to calculate by the usual rules, but the formulas will be different for positive and negative \(h .\) For \(h \neq 0,\) as \(q\) is even, \(q^{\prime}\) is odd, so \(q^{\prime \prime}\) is even, etc. and in general \(q^{(n)}(h)=(-1)^{n} q^{(n)}(-h) .\) Therefore, just as for part (b), it suffices to show that \(\lim_{h \rightarrow 0^{+}} \frac{q^{(n)}(h)}{h}=0 .\) By another induction, we see easily that for \(h>0, q^{(n)}(h)=p(1 / h) e^{-1 / h}\) for some polynomial \(p .\left[\right.\) Proof: \(q^{\prime}(h)=\left(\frac{1}{h^{2}}\right) e^{-1 / h} .\) Assuming the result for \(n,\) we have \(q^{(n+1)}(h)=\left[p(1 / h) e^{-1 / h}\right]^{\prime}=-\frac{1}{h^{2}}\left(-p(1 / h)+p^{\prime}(1 / h)\right) e^{-1 / h},\) which has the desired form.

    So we have \(\lim_{h \rightarrow 0^{+}} \frac{q^{(n)}(h)}{h}=\lim_{h \rightarrow 0^{+}} \frac{p(1 / h) e^{-1 / h}}{h}=\lim_{t \rightarrow \infty} t p(t) e^{-t}=\lim_{t \rightarrow \infty} \frac{t p(t)}{e^{t}}=0\) by part (a). Here we have again substituted \(t=1 / h\)