Course:MATH110/Archive/2010-2011/003/Notes/How to compute limits at infinity

How to compute limits at infinity?

For the moment, we don't want to discuss functions involving trigonometric, logarithmic or exponential expressions.

There are two easy examples that we actually covered in detail:

• $\displaystyle {\lim _{x\rightarrow +\infty }x}=+\infty$
• $\displaystyle {\lim _{x\rightarrow +\infty }{\frac {1}{x}}}=0$

While discussing them, we saw how using the precise definition of limits we could make the function $f(x)=x$ as large as we wanted granted $x$ was large enough and concluded that at infinity, that function would be infinity itself. For the second function, we saw how to make the function $g(x)=1/x$ as close as we want to 0, granted again that we let $x$ be large enough. This lead us to conclude that at infinity, that function tends to zero.

Strong with that understanding, it becomes fairly clear that questions such as:

• $\displaystyle {\lim _{x\rightarrow +\infty }x^{2}}=+\infty$
• $\displaystyle {\lim _{x\rightarrow +\infty }{\frac {1}{x^{2}}}}=0$

should proceed from similar arguments.

The cases that we don't know how to solve immediately are those where we get indeterminate forms, things that look like:

${\frac {0}{0}},\quad {\frac {\infty }{\infty }},\quad 0\cdot \infty ,\quad 0^{0},\quad \infty ^{\infty }$

For example, consider the following limits:

1. $\displaystyle {\lim _{x\rightarrow +\infty }{\frac {2x^{2}+3x-5}{5x^{2}+12}}}$
2. $\displaystyle {\lim _{x\rightarrow -\infty }{\frac {2x^{2}+3x-5}{5x^{2}+12}}}$
3. $\displaystyle {\lim _{x\rightarrow -\infty }{\frac {7x^{17}-3x^{12}+4x^{3}+2}{18x^{17}+4x^{6}-x}}}$
4. $\displaystyle {\lim _{t\rightarrow +\infty }{\frac {t^{7}+4t^{3}}{t^{2}+41}}}$
5. $\displaystyle {\lim _{a\rightarrow -\infty }{\frac {3a^{3}-2a}{a^{12}+73}}}$
6. $\displaystyle {\lim _{t\rightarrow +\infty }{\frac {\sqrt {t-4}}{t+2}}}$

In each of these, we can see that we have a case of infinity divided by infinity.

We can actually compute all of these with a single idea: factorize on both numerator and denominator by the largest term. This then allows for a simplification which then lets us finish the computation. In each of these cases, we obtain:

1. We can see how by doing this trick, we get rid of the indeterminate case and can finish the computation.

$\displaystyle {\lim _{x\rightarrow +\infty }{\frac {2x^{2}+3x-5}{5x^{2}+12}}=\lim _{x\rightarrow +\infty }{\frac {x^{2}(2+{\frac {3}{x}}-{\frac {5}{x^{2}}})}{x^{2}(5+{\frac {12}{x^{2}}})}}}$ $\displaystyle {=\lim _{x\rightarrow +\infty }{\frac {2+{\frac {3}{x}}-{\frac {5}{x^{2}}}}{5+{\frac {12}{x^{2}}}}}}$ $={\frac {2+0-0}{5+0}}={\frac {2}{5}}$

2. In this case, the fact that the limit is at negative infinity doesn't change anything:

$\displaystyle {\lim _{x\rightarrow -\infty }{\frac {2x^{2}+3x-5}{5x^{2}+12}}={\frac {2}{5}}}$

3. The fact that the highest degree is even or odd clearly doesn't have any influence here.

$\lim _{x\rightarrow -\infty }{\frac {7x^{17}-3x^{12}+4x^{3}+2}{28x^{17}+4x^{6}-x}}$ $=\displaystyle {\lim _{x\rightarrow -\infty }{\frac {x^{17}(7-{\frac {3}{x^{5}}}+{\frac {4}{x^{14}}}+{\frac {2}{x^{17}}})}{x^{17}(28+{\frac {4}{x^{11}}}-{\frac {1}{x^{16}}})}}}$ $=\displaystyle {\lim _{x\rightarrow -\infty }{\frac {7-{\frac {3}{x^{5}}}+{\frac {4}{x^{14}}}+{\frac {2}{x^{17}}}}{28+{\frac {4}{x^{11}}}-{\frac {1}{x^{16}}}}}}$ $={\frac {7-0+0+0}{28+0-0}}={\frac {1}{4}}$

4. When the degree of the numerator is higher than the degree of the denominator we see that the top part of the fraction wins the limit and the whole thing tends to infinity.

$\lim _{t\rightarrow +\infty }{\frac {t^{7}+4t^{3}}{t^{2}+41}}=\lim _{t\rightarrow +\infty }{\frac {t^{7}(1+{\frac {4}{t^{4}}})}{t^{2}(1+{\frac {41}{t^{7}}})}}$ $=\lim _{t\rightarrow +\infty }t^{5}{\frac {1+0}{1+0}}=+\infty \cdot 1=+\infty$

5. When the degree of the numerator is lower than the degree of the denominator, it's the opposite: the bottom part of the fraction wins and the whole thing tends to zero.

$\lim _{a\rightarrow -\infty }{\frac {3a^{3}-2a}{a^{12}+73}}=\lim _{a\rightarrow -\infty }{\frac {a^{3}(3-{\frac {2}{a^{2}}})}{a^{12}(1+{\frac {73}{a^{12}}})}}$ $=\lim _{a\rightarrow -\infty }{\frac {3-{\frac {2}{a^{2}}}}{a^{9}(1+{\frac {73}{a^{12}}})}}={\frac {3}{-\infty }}=0$

6. Even with square roots, we can still do the same thing. If you think of it, square roots are just slightly stranger exponents, so it should work fine.

$\lim _{t\rightarrow +\infty }{\frac {\sqrt {t-4}}{t+2}}=\lim _{t\rightarrow +\infty }{\frac {{\sqrt {t(1-{\frac {4}{t}}}})}{t(1+{\frac {2}{t}})}}$ $=\lim _{t\rightarrow +\infty }{\frac {t^{1/2}(1-{\frac {4}{t}})^{1/2}}{t(1+{\frac {2}{t}})}}=\lim _{t\rightarrow +\infty }{\frac {(1-{\frac {4}{t}})^{1/2}}{t^{1/2}(1+{\frac {2}{t}})}}={\frac {(1-0)^{1/2}}{\infty \cdot (1+0)}}=0$

For a limit such as:

$\displaystyle {\lim _{t\rightarrow -\infty }{\frac {\sqrt {t-4}}{t+2}}}$

It is simply not defined anywhere below 4, hence it is meaningless to try to compute its limit at negative infinity.