# Find the limit. Limit as x approaches 3^- of (x - 4)/(x - 3).

## Question:

Find the limit.

{eq}\displaystyle \lim_{x\rightarrow 3^-} \frac{x - 4}{x - 3} {/eq}

## Infinite Limits:

Suppose the limit {eq}\displaystyle\lim_{x \rightarrow c} f(x) {/eq} is equivalent to either {eq}\infty {/eq} or {eq}- \infty {/eq}.

Then, we call this limit an infinite limit.

It implies that its limit does not exist.

One particular form of an infinite limit is given by:

{eq}\displaystyle\lim_{x \rightarrow b^-} \frac{c}{x-b} = - \infty {/eq}, where {eq}c {/eq} is a positive constant

## Answer and Explanation:

It can be shown that the limit {eq}\displaystyle \lim_{x\rightarrow 3^-} \frac{x - 4}{x - 3} {/eq} is an infinite limit:

{eq}\begin{align*} \displaystyle \lim_{x\rightarrow 3^-} \frac{x - 4}{x - 3} & =\displaystyle \lim_{x\rightarrow 3^-} (x-4)\left(\frac{1}{x - 3}\right) \\ & =\displaystyle \lim_{x\rightarrow 3^-} (x-4)\displaystyle \lim_{x\rightarrow 3^-} \left(\frac{1}{x - 3}\right) \\ & =(3-4)(- \infty) \;\;\; \left[ \mathrm{Apply \ the \ infinite \ limit \ }\displaystyle\lim_{x \rightarrow b^-} \frac{c}{x-b} = - \infty, \ \mathrm{c \ is \ a \ positive \ constant}\right]\\ & = (-1) (-\infty) \\ \implies \displaystyle \lim_{x\rightarrow 3^-} \frac{x - 4}{x - 3}& = \infty\\ \end{align*} {/eq}

Hence, the limit {eq}\displaystyle \lim_{x\rightarrow 3^-} \frac{x - 4}{x - 3} {/eq} does not exist since it is an infinite limit.

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