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Limits

Limits when direct substitution cannot be used

  • An example of this is when you have a denominator, which would be equal to \(0\), therefore, the function is not continuous.
  • For example \( \displaystyle \lim_{x\rightarrow 1} \left(\frac{x^{2}-1}{x-1}\right) \). When \(x=1\), the function is undefined, as the denominator would equal \(0\).
  • When this happens, try to re-write the function as something simpler. For example:
    \begin{eqnarray*}  
    \lim_{x\rightarrow 1} \left(\frac{x^{2}-1}{x-1}\right)  
    &=& \lim_{x\rightarrow 1} \left(\frac{(x+1)(x-1)}{x-1}\right) \\  
    &=& \lim_{x\rightarrow 1} \left(x+1\right) \\  
    &=& 1+1 \\  
    &=& 2 
    \end{eqnarray*}

  • Another example:
    \begin{eqnarray*}  
    \lim_{h\rightarrow 0} \frac{(4+h)^2-16}{h}  
    &=& \lim_{h\rightarrow 0}\frac{16 +8h+h^2 - 16}{h} \\  
    &=& \lim_{h\rightarrow 0}\frac{8h + h^2}{h} \\  
    &=& \lim_{h\rightarrow 0}\frac{\cancel{h} (8 +h)}{\cancel{h}} \\  
    &=& \lim_{h\rightarrow 0} 8 +h\\  
    &=& 8
    \end{eqnarray*}

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