# Math Help - Help needed in Limits

1. ## Help needed in Limits

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. 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

3. 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

4. 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 $x \to a$.

5. 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 $x \to \sqrt{2}$.

6. 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

7. 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 $x \to a$

8. ## hi

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