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Limits and continuity · Limits

Limits at infinity x → ±∞

In 1st Bachillerato you already learned to compare the degrees of polynomials and rational functions when $x\to\pm\infty$. Now the challenge is subtler: comparing infinities of different kinds (an exponential against a polynomial, a logarithm against a root) and resolving indeterminate forms that can no longer be sorted out just by looking at degrees.

Before you start, make sure you know: Limit of a function as x → ∞ (1st Bachillerato)
Before you start, make sure you know: Calculating limits of functions (1st Bachillerato)

Revision: polynomials and rational functions, now with parameters

In university entrance exams (EBAU) these rules almost always come with an unknown parameter that you have to find from a condition on the limit.

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Determine the value of the following limits based on comparing the degrees of the polynomials: a) $\lim_{x \to +\infty} \frac{5x^2 - 3x + 8}{2x^2 + 1}$ b)…

Determine the value of the following limits based on comparing the degrees of the polynomials:

  1. a) $\lim_{x \to +\infty} \frac{5x^2 - 3x + 8}{2x^2 + 1}$
  1. b) $\lim_{x \to -\infty} \frac{x^3 + 2x}{x^4 - 5}$
Solve
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Given the rational function $f(x) = \dfrac{ax^2 + 3x - 5}{2x^2 - k}$, calculate the values of the real parameters $a$ and $k$ knowing that the function has a horizontal…

Given the rational function $f(x) = \dfrac{ax^2 + 3x - 5}{2x^2 - k}$, calculate the values of the real parameters $a$ and $k$ knowing that the function has a horizontal asymptote at $y = 3$ and that its graph passes through the point $(1, -2)$. Once found, check whether the limit as $x \to -\infty$ coincides with the limit as $x \to +\infty$.

x y 0 1 2 −1 −2 1 −2 y = 3 x = -√2 x = √2 (1, −2) x → -∞ x → +∞
Solve

Hierarchy of infinities: exponential, polynomial and logarithm

When $x\to+\infty$, not all "infinities" grow at the same rate. There is a fixed order worth memorising: if you divide the one that grows faster by the one that grows more slowly, the limit is $+\infty$; the other way round, it is $0$.

Graph comparing the growth towards infinity of the natural logarithm, x to the power of ten and two to the power of x. ln(x) x¹⁰ 2ˣ ln(x) << x¹⁰ << 2ˣ when x is very large
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Given the functions $f(x) = 2^x$ and $g(x) = x^{10} + 500x^2$, reason which of the two functions grows faster as $x \to +\infty$ using the scale of infinities. From your…

Given the functions $f(x) = 2^x$ and $g(x) = x^{10} + 500x^2$, reason which of the two functions grows faster as $x \to +\infty$ using the scale of infinities. From your answer, deduce the value of the following limit:

$$\lim_{x \to +\infty} \frac{x^{10} + 500x^2}{2^x}$$
Solve
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Given the functions $f(x) = e^x$, $g(x) = x^{10}$, and $h(x) = \ln(x)$, order the following limits from smallest to largest according to the order of magnitude of the…

Given the functions $f(x) = e^x$, $g(x) = x^{10}$, and $h(x) = \ln(x)$, order the following limits from smallest to largest according to the order of magnitude of the infinities involved as $x \to +\infty$:

  1. $L_1 = \lim_{x \to +\infty} \dfrac{g(x)}{f(x)}$

  2. $L_2 = \lim_{x \to +\infty} \dfrac{g(x)}{h(x)}$

  3. $L_3 = \lim_{x \to +\infty} \dfrac{h(x)}{f(x)}$

Solve

The number $e$ as a limit: the $1^\infty$ indeterminate form

There is a special limit that appears again and again when studying growth (compound interest, populations, decay): a base that tends to $1$ raised to an exponent that tends to $\infty$. This form, $1^\infty$, is indeterminate and is resolved using the number $e$.

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Complete the following table of values for the function $f(x) = \left( \frac{x+1}{x} \right)^x$ and use the results obtained to conjecture which famous number the…

Complete the following table of values for the function $f(x) = \left( \frac{x+1}{x} \right)^x$ and use the results obtained to conjecture which famous number the function tends to as $x$ becomes very large.

$x$ $10$ $100$ $1000$ $10000$
$f(x)$ $2,5937$ $2,7048$ $2,7169$ $2,7181$

Write the result using limit notation:

$$\lim_{x \to +\infty} \left( 1 + \frac{1}{x} \right)^x = \dots$$
Solve
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Determine the value of the following limit using its relation to the number $e$. You must transform the expression so that it takes the form of the definition of this…

Determine the value of the following limit using its relation to the number $e$. You must transform the expression so that it takes the form of the definition of this constant:

$$\lim_{x \to +\infty} \left( \frac{2x + 5}{2x - 1} \right)^{x+3}$$
Solve

Roots: the conjugate technique for $\infty-\infty$

When you get a subtraction of roots that individually both tend to $+\infty$ or both tend to $-\infty$, you cannot simply subtract: you have to multiply and divide by the conjugate to turn the subtraction into a quotient.

Diagram of the conjugate technique: multiplying and dividing √A − √B by √A + √B turns it into the quotient (A − B) / (√A + √B). √A − √B × (√A + √B) ÷ (√A + √B) A − B √A + √B
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Solve the following limit involving a difference of irrational functions. Remember to use the conjugate method to resolve the indeterminate form $[\infty - \infty]$:…

Solve the following limit involving a difference of irrational functions. Remember to use the conjugate method to resolve the indeterminate form $[\infty - \infty]$:

$$\lim_{x \to +\infty} \left( \sqrt{x^2 + 5x} - \sqrt{x^2 - 3x} \right)$$
Solve
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Study the limit at negative infinity of the following rational function containing an absolute value and a square root: $$f(x) = \frac{|3x - 2| + \sqrt{x^2 + 5x}}{2x -…

Study the limit at negative infinity of the following rational function containing an absolute value and a square root:

$$f(x) = \frac{|3x - 2| + \sqrt{x^2 + 5x}}{2x - 7}$$

That is, calculate $\lim_{x \to -\infty} f(x)$. Take special care with the sign of the variable $x$ when inserting or extracting factors from the square root and when removing the absolute value.

Solve
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Calculate the limit of the sequence of functions given by the following nested root: $$\lim_{x \to +\infty} \left( \sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x} \right)$$…

Calculate the limit of the sequence of functions given by the following nested root:

$$\lim_{x \to +\infty} \left( \sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x} \right)$$

This exercise requires a precise use of the conjugate technique and a careful analysis of the dominant power within the radicals.

Solve

Back to the start: To compare a logarithm against a root such as $\lim_{x\to+\infty}\frac{\ln(x)}{\sqrt{x}}$, write the root as the power $x^{0{,}5}$; by the scale of infinities ($n=0{,}5>0$), the power outgrows the logarithm and the limit is $0$.

Summary

See all the exercises in this section