Link Search Menu Expand Document
CMI Entrance Exam

Limits

Vanilla application of L’Hospital

A3, 2010

Evaluate the limit:

Solution We apply the L'Hospital's rule and differentiate both the numerator and the denominator.

Practice Problem

Explain what is wrong with the use of L'Hospital's rule. Find the correct limit.

L’Hospital again

A4, 2012

Prove the following limit.

Solution

For some positive constant we can ensure that for any . For example, will work. This is because and is an increasing function.

Further, for . So the given ratio must lie between 0 and . 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 as defined below:

  1. As , the sign of changes infinitely many times.
  2. does not exist.
  3. .
  4. .

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

    • Tower expression

    A7, 2022

    Let ,,, etc. For which of the following options are true?

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

    • Limit of

    B1a, 2017

    (a) Evaluate

    Solution

    First consider the limit

    Now consider just the exponent: substituting the value 0 from (2) in equation (1) we get that the limit is 1.

    Now

    Reference. This problem is based on a general result: A tower of even number of s tends to zero and a tower of odd number of 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 and be function defined as follows:
    (a)
    (b) does not exist.
    (c) is finite.
    (d) .

    Solution (a) True. as .
    (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 be a continuous function from to (where is the set of all real numbers) that satisfies the following property:

    For every natural number

    For example, Calculate the following.

    (a)

    (b) The number of solutions to the equation

    Solution

    (a) will take value 2 for all even . At the same time, primes provide an increasing infinite sequence of positive integers for which Thus does not exist.

    (b) By intermediate value theorem, for each prime there is an between and such that

    Limit of a log of an exponent

    A9, 2019

    Consider defined as follows:

    (a) is not onto i.e. the range of is not all of .

    (b) For every the function is continuous everywhere.

    (c) For every the function is differentiable everywhere.

    (d) We have for all

    Solution

    Without loss of generality, assume that .

    Similary, if , . Thus, we have:

    1. False. so is onto.
    2. True. which is continuous everywhere.
    3. False. is not differentiable at . More explicitly, and .
    4. True. for all .

    Polynomial and limits

    B1, 2015

    (a) For any polynomial the limit is independent of the polynomial Justify this statement and find the value of the limit.

    (b) Consider the function defined by

    Show that exists and find its value. Why is it enough to calculate the relevant limit from only one side?

    (c) Now for any positive integer show that exists and find its value. Here is the function in part (b) and denotes its -th derivative at .

    Solution

    (a) If is constant, then the limit Otherwise we get a form . Using L'Hospital's rule, we get by induction on the degree of (or apply L'Hospital's rule repeatedly).


    (b) The right side derivative (Let As This limit is e.g. by part Now is an even function, so letting the left side derivative Using the earlier calculation this also equals Note: It is wrong to argue that because to do so, we first need to know that is continuous at but we have not even shown that exists! For the same reason it is wrong to argue below that


    (c) We will show by induction on that The case is done. (We can even start the induction at by interpreting ) Assuming that we are done up to and to prove the statement for we need to calculate because by induction. Therefore it is good to examine for This is easy to calculate by the usual rules, but the formulas will be different for positive and negative For as is even, is odd, so is even, etc. and in general Therefore, just as for part (b), it suffices to show that By another induction, we see easily that for for some polynomial Proof: Assuming the result for we have which has the desired form.

    So we have by part (a). Here we have again substituted