# Evaluate the limit \lim_{h\rightarrow 0 } \frac{(7+h)^3-343}{h}

## Question:

Evaluate the limit {eq}\displaystyle \lim_{h\rightarrow 0 } \frac{(7+h)^3-343}{h} {/eq}

## L'Hospital's Rule:

The forms {eq}\displaystyle \frac{\infty}{\infty} {/eq} and {eq}\displaystyle \frac{0}{0} {/eq} are called indeterminate forms.

If a limit follows either of these forms, then L'Hospital's rule should be applied.

This particular rule states that:

{eq}\displaystyle\lim_{x \rightarrow b}\frac{f(x)}{g(x)}=\displaystyle\lim_{x \rightarrow b}\frac{f'(x)}{g'(x)} {/eq}, provided the limit {eq}\displaystyle\lim_{x \rightarrow b}\frac{f(x)}{g(x)} {/eq} contains either of the stated forms.

## Answer and Explanation:

Apply L'Hospital's rule to the limit {eq}\displaystyle \lim_{h\rightarrow 0 } \frac{(7+h)^3-343}{h} {/eq} since it has the form {eq}\displaystyle \frac{0}{0} {/eq}, which is an indeterminate form:

{eq}\begin{align*} \displaystyle \lim_{h\rightarrow 0 } \frac{(7+h)^3-343}{h} & = \displaystyle \lim_{h\rightarrow 0 } \frac{D_h((7+h)^3-343)}{D_h(h)} \;\;\; \left[ \mathrm{ L'Hospital's \ Rule }\right]\\ & = \displaystyle \lim_{h\rightarrow 0 } \frac{3( 7 + h)^ 2 }{1}\\ & = \frac{3( 7 + 0)^ 2 }{1}\\ \implies \displaystyle \lim_{h\rightarrow 0 } \frac{(7+h)^3-343}{h}& = 147\\ \end{align*} {/eq}

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