Question

Suppose that we know the reciprocal rule for limits: If $\underset{\mathbf{x}\mathbf{\to }\mathbf{c}}{\mathbf{lim}}\mathit{g}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{M}$exists and is nonzero, then $\underset{\mathbf{x}\mathbf{\to }\mathbf{c}}{\mathbf{lim}}\frac{\mathbf{1}}{\mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{M}}$This limit rule is tedious to prove and we do not include it here. Use the reciprocal rule and the product rule for limits to prove the quotient rule for limits.

Short answer

Ans: $\underset{\mathbf{x}\mathbf{\to }\mathbf{c}}{\mathbf{lim}}\frac{\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}=\frac{{\mathbf{lim}}_{\mathbf{x}\mathbf{\to }\mathbf{c}}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{{\mathbf{lim}}_{\mathbf{x}\mathbf{\to }\mathbf{c}}\mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}$

Step-by-step solution

Step 1. Given Information:

The reciprocal rule is given by,

if $\underset{\mathbf{x}\mathbf{\to }\mathbf{c}}{\mathbf{lim}}\mathit{g}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{M}$exists and is nonzero, then$\underset{\mathbf{x}\mathbf{\to }\mathbf{c}}{\mathbf{lim}}\frac{\mathbf{1}}{\mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{M}}$

Step 2. Prove:

If $\underset{\mathbf{x}\mathbf{\to }\mathbf{c}}{\mathbf{lim}}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}$exists and $\underset{\mathbf{x}\mathbf{\to }\mathbf{c}}{\mathbf{lim}}\mathit{g}\mathbf{\left(}\mathit{x}\mathbf{\right)}$exists and is nonzero, then by the product and reciprocal rules for limit we have,

localid="1668094027457" $$\begin{gathered} \underset{\mathbf{x}\mathbf{\to }\mathbf{c}}{\mathbf{l}\mathbf{i}\mathbf{m}}\frac{\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}\mathbf{=}\underset{\mathbf{x}\mathbf{\to }\mathbf{c}}{\mathbf{l}\mathbf{i}\mathbf{m}}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\frac{\mathbf{1}}{\mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}} \\ \mathbf{}\mathbf{}\mathbf{=}\underset{\mathbf{x}\mathbf{\to }\mathbf{c}}{\mathbf{l}\mathbf{i}\mathbf{m}}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\underset{\mathbf{x}\mathbf{\to }\mathbf{c}}{\mathbf{l}\mathbf{i}\mathbf{m}}\frac{\mathbf{1}}{\mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}} \\ \mathbf{}\mathbf{}\mathbf{=}\underset{\mathbf{x}\mathbf{\to }\mathbf{c}}{\mathbf{l}\mathbf{i}\mathbf{m}}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\frac{\mathbf{1}}{\mathbf{l}\mathbf{i}{\mathbf{m}}_{\mathbf{x}\mathbf{\to }\mathbf{c}}\mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}} \\ \mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{l}\mathbf{i}{\mathbf{m}}_{\mathbf{x}\mathbf{\to }\mathbf{c}}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathbf{l}\mathbf{i}{\mathbf{m}}_{\mathbf{x}\mathbf{\to }\mathbf{c}}\mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}} \end{gathered}$$