How do you find the limit of $\frac{x + 3}{x - 3}$ $(x+3)/(x-3)$ as x approaches 3+?

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Answer 1 · 2015-04-22T17:50:09

As $x \to 3^{+}$ $xrarr3^+$ , the numerator $x + 3$ $x+3$ $\to 6$ $rarr 6$ .

The denominator $x - 3$ $x-3$ $\to 0$ $rarr 0$ .

So the fraction is either increasing without bound ( $\to \infty$ $rarroo$ ) or decreasing without bound ( $\to - \infty$ $rarr -oo$ )

As $x \to 3^{+}$ $xrarr3^+$ , we know that $x > 3$ $x > 3$ so $x - 3 > 3 - 3 = 0$ $x-3 > 3-3 =0$ . That is

As $x \to 3^{+}$ $xrarr3^+$ , $x - 3$ $x-3$ $\to 0$ $rarr 0$ through positive values.

(As $x \to 3^{+}$ $xrarr3^+$ , $x - 3$ $x-3$ $\to 0^{+}$ $rarr 0^+$ or $lim_(xrarr3^+)(x-3) = 0^+$ )

The numerator is going to a positive and the denominator is going to 0, through positives, so the fraction is increasing without bound.

$lim_(xrarr3^+)(x+3)/(x-3) = oo$

How do you find the limit of (x+3)/(x-3)x+3x−3(x+3)/(x-3) as x approaches 3+?