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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*}