Course:MATH110/Archive/2010-2011/003/Notes/Indeterminate forms

A lot of limits can be computed by simply plugging-in the value. This is due to the continuity of the function and is explained in this note. There are times where such a thing leads to expressions of the type:

${\frac {0}{0}}$    or    ${\frac {\infty }{\infty }}$    or    $0\cdot \infty$    or    $\displaystyle {\infty -\infty }$

Those are called indeterminate forms and you can not conclude immediately the value of the limit. More work is required is these cases since as the examples below show, the limit can end up being anything.

Examples

For the case $\infty -\infty$

a) $\lim _{x\rightarrow 1}\left({\frac {1}{x^{2}-x}}-{\frac {2}{x^{2}-1}}\right)$

b) $\lim _{x\rightarrow +\infty }\left(x^{2}-x\right)$

Both of these examples showcase a limit that appears of the indeterminate form $\infty -\infty$, but as we'll see below, they do not yield the same result.

Solutions

a) $\lim _{x\rightarrow 1}\left({\frac {1}{x^{2}-x}}-{\frac {2}{x^{2}-1}}\right)$

$=\lim _{x\rightarrow 1}\left({\frac {1}{x(x-1)}}-{\frac {2}{(x+1)(x-1)}}\right)$

$=\lim _{x\rightarrow 1}\left({\frac {(x+1)}{x(x-1)(x+1)}}-{\frac {2x}{x(x-1)(x+1)}}\right)$

$=\lim _{x\rightarrow 1}{\frac {x+1-2x}{x(x-1)(x+1)}}$

$=\lim _{x\rightarrow 1}{\frac {-(x-1)}{x(x-1)(x+1)}}$

$=\lim _{x\rightarrow 1}{\frac {-1}{x(x+1)}}=-{\frac {1}{2}}$

b) $\lim _{x\rightarrow +\infty }\left(x^{2}-x\right)$

$=\lim _{x\rightarrow +\infty }\left(x^{2}(1-{\frac {1}{x}})\right)$

$=\left(\lim _{x\rightarrow +\infty }x^{2}\right)\left(\lim _{x\rightarrow +\infty }1-{\frac {1}{x}}\right)$

$=\left(\lim _{x\rightarrow +\infty }x^{2}\right)\cdot 1$

$=\lim _{x\rightarrow +\infty }x^{2}=+\infty$