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*}