# [SOLVED] Limit Problem

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• May 26th 2008, 12:16 AM
chukie
[SOLVED] Limit Problem
Hi I was wondering if someone could help me with this limit problem:

Find the lim x->a-, lim x->a+ and lim x->a
for the f(x)=(x^2-16)/(squareroot(x^2-8x+16)) a=4

I've tried to simplify the problem but after that i dont know wut to do =(
This is wut I've done:
y=(x^2-16)/(squareroot(x-4)^2)
y=(x^2-16)/abs(x-4)

any help would be appreciated.
• May 26th 2008, 12:23 AM
Moo
Hello,

Quote:

Originally Posted by chukie
Hi I was wondering if someone could help me with this limit problem:

Find the lim x->a-, lim x->a+ and lim x->a
for the f(x)=(x^2-16)/(squareroot(x^2-8x+16)) a=4

I've tried to simplify the problem but after that i dont know wut to do =(
This is wut I've done:
y=(x^2-16)/(squareroot(x-4)^2)
y=(x^2-16)/abs(x-4)

any help would be appreciated.

When x->4-, x-4 is negative. Therefore abs(x-4)=-(x-4).
Then, simplify with x²-16=(x-4)(x+4)

When x->4+, x-4 > 0. Therefore abs(x-4)=(x-4).
• May 26th 2008, 12:29 AM
Reckoner
Quote:

Originally Posted by chukie
Hi I was wondering if someone could help me with this limit problem:

Find the lim x->a-, lim x->a+ and lim x->a
for the f(x)=(x^2-16)/(squareroot(x^2-8x+16)) a=4

I've tried to simplify the problem but after that i dont know wut to do =(
This is wut I've done:
y=(x^2-16)/(squareroot(x-4)^2)
y=(x^2-16)/abs(x-4)

any help would be appreciated.

The numerator is a difference of squares, so you should be able to factor it:

$\displaystyle \frac{x^2-16}{\sqrt{x^2 - 8x + 16}}$

$\displaystyle =\frac{\left(x + 4\right)\left(x - 4\right)}{\sqrt{\left(x - 4\right)^2}}$

$\displaystyle =\left(x + 4\right)\,\frac{x-4}{|x - 4|}$

Now, when you take the limit, notice that $\displaystyle x - 4$ is always negative when $\displaystyle x<4$ (which will be the case for the left-handed limit) and nonnegative when $\displaystyle x\geq4$ (which will be the case for the right-handed limit). So, get rid of the absolute value bars (negating when appropriate) and simplify.
• May 26th 2008, 09:48 AM
chukie
thanks for ur help in the previous question. i was also wondering if the absolute value function was in the numerator, wut sign would i need to take for the lim->a-, lim ->a+?
Example:
f(x)=abs(x^2+5x+6)/(x^2-9)

i did the same thing as u guys hv instructed me for the other question by taking a negative value of the absolute value of the numerator wen lim->a- and positive wen lim->a+. then i was able to simplify. i was able to get a number for limits from the left and from the right. but wen i graph the function on my calculator i see that the answer is supposed to be negative infinity for the left and positive infinity for the right. im a bit confused :confused:
• May 26th 2008, 10:01 AM
Reckoner
Quote:

Originally Posted by chukie
thanks for ur help in the previous question. i was also wondering if the absolute value function was in the numerator, wut sign would i need to take for the lim->a-, lim ->a+?
Example:
f(x)=abs(x^2+5x+6)/(x^2-9)

$\displaystyle \frac{|x^2+5x+6|}{x^2-9}$

$\displaystyle = \frac{|(x + 2)(x + 3)|}{(x + 3)(x - 3)}$

$\displaystyle = \frac{|x + 2||x + 3|}{(x + 3)(x - 3)}$

Now, the absolute value will depend on the values that $\displaystyle x$ is taking. As $\displaystyle x\to a-$, is $\displaystyle x + 3$ positive or negative? What about $\displaystyle x + 2$ and $\displaystyle x - 3$? How about as $\displaystyle x\to a+$? This will all, of course, depend on the value of $\displaystyle a$.

When the expression is positive, you can remove the absolute bars, and when it is negative you can remove the bars as long as you negate the expression.
• May 26th 2008, 06:50 PM
chukie
Quote:

Originally Posted by Reckoner
$\displaystyle \frac{|x^2+5x+6|}{x^2-9}$

$\displaystyle = \frac{|(x + 2)(x + 3)|}{(x + 3)(x - 3)}$

$\displaystyle = \frac{|x + 2||x + 3|}{(x + 3)(x - 3)}$

Now, the absolute value will depend on the values that $\displaystyle x$ is taking. As $\displaystyle x\to a-$, is $\displaystyle x + 3$ positive or negative? What about $\displaystyle x + 2$ and $\displaystyle x - 3$? How about as $\displaystyle x\to a+$? This will all, of course, depend on the value of $\displaystyle a$.

When the expression is positive, you can remove the absolute bars, and when it is negative you can remove the bars as long as you negate the expression.

so if a=3
then from wut u mean wen the lim->3-, the equation will become ((x+2)(x+3))/((x+3)(x-3)) since the absolute value expression will be positive? then simplifying it will become (x+2)/(x-3)?
• May 26th 2008, 06:56 PM
Reckoner
Quote:

Originally Posted by chukie
so if a=3
then from wut u mean wen the lim->3-, the equation will become ((x+2)(x+3))/((x+3)(x-3)) since the absolute value expression will be positive? then simplifying it will become (x+2)/(x-3)?

That is correct.