Notation for Limits

We say that the limit of f as x approaches a is L and we write $\lim_{{x\rightarrow a}}^{}$f (x) = L if for each number $\varepsilon$ > 0 there exists a number δ > 0 such that $\left\vert\vphantom{ f(x)-L}\right.$f (x) - L$\left.\vphantom{ f(x)-L}\right\vert$ < ε whenever 0 < $\left\vert\vphantom{ x-a}\right.$x - a$\left.\vphantom{ x-a}\right\vert$ < δ.

$\blacktriangleright$ To find a limit of the form $\lim_{{x\rightarrow a}}^{}$f (x) using Scientific Notebook

1.
Type UserInputlim while in mathematics mode, or click itbpF0.3009in0.3009in0.0701infunction.wmfand choose lim.

2.
Click itbpF0.3009in0.3009in0.0701insubscrip.wmf and enter x→a as a subscript.

3.
Enter f (x).

4.
Choose Evaluate, press CTRL + E, or click itbpF0.3009in0.3009in0.0701inevaluate.wmf.

$\blacktriangleright$ Evaluate

$\lim\limits_{{x\rightarrow 1}}^{}$${\dfrac{{x^{2}-1}}{{x-1}}}$ = 2

This result is reasonable, since for x≠1, ${\frac{{x^2-1}}{{x-1}% }}$ = x + 1, which is close to 2 when x is close to 1.

Limits of rational functions are not always apparent. You cannot evaluate the following expression at x = - 3/2, because the denominator is 0 at this point. However the expression has a limit at -3/2.

$\blacktriangleright$ Evaluate

$\lim\limits_{{x\rightarrow -3/2}}^{}$${\dfrac{{4x^{4}+6x^{2}+19x+6x^{3}+15}}{{% 2x^{3}+5x^{2}+5x+3}}}$ = - ${\dfrac{{25}}{{7}}}$

Factor in place and Simplify in place allow you to fill in the steps leading to evaluation by direct substitution, because the second step removes the singularity from the expression. For the first two lines, copy the entire expression after an equals sign, and carry out in-place operations. For the third line, substitute x = - 3/2 into the expression.

$\blacktriangleright$ Factor, Simplify, ..., Evaluate


$\displaystyle \lim_{{x\rightarrow -3/2}}^{}$$\displaystyle {\frac{{4x^{4}+6x^{2}+19x+6x^{3}+15}}{{2x^{3}+5x^{2}+5x+3}}}$ = $\displaystyle \lim_{{x\rightarrow -3/2}}^{}$$\displaystyle {\frac{{\left( 2x+3\right) \left( x+1\right) \left( 2x^{2}-2x+5\right) }}{{\left( 2x+3\right) \left( x^{2}+x+1\right) }% }}$ 6pt  
  = $\displaystyle \lim_{{x\rightarrow -3/2}}^{}$$\displaystyle \left(\vphantom{ x+1}\right.$x + 1$\displaystyle \left.\vphantom{ x+1}\right)$$\displaystyle {\frac{{2x^{2}-2x+5}}{{x^{2}+x+1}% }}$ 6pt  
  = $\displaystyle \left[\vphantom{ \left( x+1\right) \frac{2x^{2}-2x+5}{x^{2}+x+1}}\right.$$\displaystyle \left(\vphantom{ x+1}\right.$x + 1$\displaystyle \left.\vphantom{ x+1}\right)$$\displaystyle {\frac{{2x^{2}-2x+5}}{{x^{2}+x+1}}}$$\displaystyle \left.\vphantom{ \left( x+1\right) \frac{2x^{2}-2x+5}{x^{2}+x+1}}\right]_{{x=-3/2}% }^{}$ 6pt  
  = - $\displaystyle {\frac{{25}}{{7}}}$  
       

You can carry out the substitutionDM3-4.tex#Substitution in the third line of the preceding example as follows.

$\blacktriangleright$ To substitute a value into the expression

1.
Select the expression $\left(\vphantom{ x+1}\right.$x + 1$\left.\vphantom{ x+1}\right)$${\frac{{2x^{2}-2x+5}}{{% x^{2}+x+1}}}$ with the mouse or SHIFT + RIGHT ARROW.

2.
Click itbpF0.3009in0.3009in0.0701insqbrack.wmf.

3.
Click itbpF0.3009in0.3009in0.0701insubscrip.wmfand enter x = - 3/2 as a subscript.

4.
Choose Evaluate.


$\blacktriangleright$ Evaluate

$\left[\vphantom{ \left( x+1\right) \frac{2x^{2}-2x+5}{x^{2}+x+1}}\right.$$\left(\vphantom{ x+1}\right.$x + 1$\left.\vphantom{ x+1}\right)$${\frac{{2x^{2}-2x+5}}{{x^{2}+x+1}}}$$\left.\vphantom{ \left( x+1\right) \frac{2x^{2}-2x+5}{x^{2}+x+1}}\right]_{{x=-3/2}}^{}$ = - ${\frac{{25}}{{7}}}$

You can also carry out the replacement in the third line using the editing features of Scientific Notebook.

$\blacktriangleright$ To do an automatic replacement of mathematics

1.
Select the expression $\left(\vphantom{ x+1}\right.$x + 1$\left.\vphantom{ x+1}\right)$${\frac{{2x^{2}-2x+5}}{{% x^{2}+x+1}}}$ with the mouse or SHIFT + RIGHT ARROW.

2.
From the Edit menu, choose Replace.

3.
Fill in the choices in the dialog box in mathematics mode.

The result is the expression

$\displaystyle \left(\vphantom{ \left( -3/2\right) +1}\right.$$\displaystyle \left(\vphantom{ -3/2}\right.$ -3/2$\displaystyle \left.\vphantom{ -3/2}\right)$ + 1$\displaystyle \left.\vphantom{ \left( -3/2\right) +1}\right)$$\displaystyle {\frac{{2\left( -3/2\right) ^{2}-2\left( -3/2\right) +5}}{{\left( -3/2\right) ^{2}+\left( -3/2\right) +1}}}$