Help needed in Limits

**MathematicalDick** · March 27th 2010, 12:06 AM

Please, kindly explain the solution step by step.

Limit problems are in $\frac{0}{0}$ form.

1. $\lim_{x\to \ -2} \frac {x^3+x^2+4x+12}{x^3-3x+2}$

2. $\lim_{x\to \ a} \frac {\sqrt{x}-\sqrt{a}}{\sqrt{x^2-a^2}}$

3. $\lim_{x\to \ \sqrt{2}} \frac {x^2-2}{x^2+\sqrt{2x}-4}$

**running-gag** · March 27th 2010, 12:33 AM

Hi

Originally Posted by MathematicalDick

1. $\lim_{x\to \ -2} \frac {x^3+x^2+4x+12}{x^3-3x+2}$

The fact that you find 0/0 means that you can factor out (x+2) to both numerator and denominator

**running-gag** · March 27th 2010, 12:35 AM

Originally Posted by MathematicalDick

2. $\lim_{x\to \ a} \frac {\sqrt{x}-\sqrt{a}}{\sqrt{x^2-a^2}}$

http://www.mathhelpforum.com/math-he...tml#post480676

**Prove It** · March 27th 2010, 12:36 AM

Originally Posted by MathematicalDick

Please, kindly explain the solution step by step.

Limit problems are in $\frac{0}{0}$ form.

1. $\lim_{x\to \ -2} \frac {x^3+x^2+4x+12}{x^3-3x+2}$

2. $\lim_{x\to \ a} \frac {\sqrt{x}-\sqrt{a}}{\sqrt{x^2-a^2}}$

3. $\lim_{x\to \ \sqrt{2}} \frac {x^2-2}{x^2+\sqrt{2x}-4}$

2. $\frac{\sqrt{x} - \sqrt{a}}{\sqrt{x^2 - a^2}} = \left(\frac{\sqrt{x} - \sqrt{a}}{\sqrt{x^2 - a^2}}\right)\left(\frac{\sqrt{x^2 - a^2}}{\sqrt{x^2 - a^2}}\right)$

$= \frac{(\sqrt{x} - \sqrt{a})\sqrt{x^2 - a^2}}{x^2 - a^2}$

$= \frac{(\sqrt{x} - \sqrt{a})\sqrt{x^2 - a^2}}{(x - a)(x + a)}$

$= \left[\frac{(\sqrt{x} - \sqrt{a})\sqrt{x^2 - a^2}}{(x - a)(x + a)}\right]\left(\frac{\sqrt{x} + \sqrt{a}}{\sqrt{x} + \sqrt{a}}\right)$

$= \frac{(x - a)\sqrt{x^2 - a^2}}{(x - a)(x + a)(\sqrt{x} + \sqrt{a})}$

$= \frac{\sqrt{x^2 - a^2}}{(x + a)(\sqrt{x} + \sqrt{a})}$ .

Now make $x \to a$ .

**Prove It** · March 27th 2010, 12:42 AM

Originally Posted by MathematicalDick

Please, kindly explain the solution step by step.

Limit problems are in $\frac{0}{0}$ form.

1. $\lim_{x\to \ -2} \frac {x^3+x^2+4x+12}{x^3-3x+2}$

2. $\lim_{x\to \ a} \frac {\sqrt{x}-\sqrt{a}}{\sqrt{x^2-a^2}}$

3. $\lim_{x\to \ \sqrt{2}} \frac {x^2-2}{x^2+\sqrt{2x}-4}$

3. $\frac{x^2 - 2}{x^2 + \sqrt{2}x - 4} = \frac{(x - \sqrt{2})(x + \sqrt{2})}{(x - \sqrt{2})(x + 2\sqrt{2})}$

$= \frac{x + \sqrt{2}}{x + 2\sqrt{2}}$ .

Now make $x \to \sqrt{2}$ .

**MathematicalDick** · March 27th 2010, 12:59 AM

Originally Posted by running-gag

http://www.mathhelpforum.com/math-he...tml#post480676

I absolutely forgot factorization. I factored $x^3-3x+2$ as $(x-1)(x^2+x-2)$ before you gave the the hint that both the Nr and the Dr will be factored by $x+2$ .

I finally got the answer $\lim_{x\to \-2} \frac{(x+2)(x^2-x+6)}{(x+2)(x^2-2x+1)}$ = $\frac{(-2)^2-(-2)+6}{(-2)^2-2(-2)+1}$ = $\frac{4}{3}$

I want to revise the factorization, may I request you to kindly send me the tutorial link of these kind of factorials in this community. Thanks

**MathematicalDick** · March 27th 2010, 01:50 AM

Originally Posted by Prove It

2. $\frac{\sqrt{x} - \sqrt{a}}{\sqrt{x^2 - a^2}} = \left(\frac{\sqrt{x} - \sqrt{a}}{\sqrt{x^2 - a^2}}\right)\left(\frac{\sqrt{x^2 - a^2}}{\sqrt{x^2 - a^2}}\right)$

$= \frac{(\sqrt{x} - \sqrt{a})\sqrt{x^2 - a^2}}{x^2 - a^2}$

$= \frac{(\sqrt{x} - \sqrt{a})\sqrt{x^2 - a^2}}{(x - a)(x + a)}$

$= \left[\frac{(\sqrt{x} - \sqrt{a})\sqrt{x^2 - a^2}}{(x - a)(x + a)}\right]\left(\frac{\sqrt{x} + \sqrt{a}}{\sqrt{x} + \sqrt{a}}\right)$

$= \frac{(x - a)\sqrt{x^2 - a^2}}{(x - a)(x + a)(\sqrt{x} + \sqrt{a})}$

$= \frac{\sqrt{x^2 - a^2}}{(x + a)(\sqrt{x} + \sqrt{a})}$ .

Now make $x \to a$ .

hmmm

Why so long? How about this?

$\lim_{x\to \ a} \left(\frac {\sqrt{x}-\sqrt{a}}{\sqrt{x^2-a^2}}\right)$ X $\left(\frac {\sqrt{x}+\sqrt{a}}{\sqrt{x}+\sqrt{a}}\right)$

= $\lim_{x\to \ a}\frac{x-a}{(\sqrt{x}+\sqrt{a})\sqrt{x^2-a^2}}$

= $\lim_{x\to \ a}\frac{\sqrt{x-a}}{(\sqrt{x}+\sqrt{a})\sqrt{x+a}}$

Now make $x \to a$

**maclhen** · April 7th 2010, 01:49 AM

this is long method

.

Now make .

this is the short cut method

X

=

....its the easy way to answer your question...

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Transparent Coupon Code

**MathematicalDick** · April 7th 2010, 04:43 AM

Thanks. But I have already got the answer and posted the same thing just above your post.

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