find the limits of the function

**williamb** · February 7th 2009, 04:35 PM

consider the function

g(x) = [((x-8)^2)*((x-1)^3)] / [sqrt(x-5)]

i need to find the limits ..
x -> 8+
x-> 8-
x-> 5+
x-> 5-
x-> 1+
x-> 1-

i've found
x -> 8+ = infinity
x-> 8- = infinity
x-> 5+ = infinity
x-> 5- = DNE
x-> 1+ = DNE
x-> 1- = DNE

i thought that the last three didn't exsist because the square root was of a negative number when these were used, i know how to find limits but obviously i'm missing something here.
i was almost certain that sqrt(-) cannot be done for this,
please help me it's driving me mad

thanks in advance,
brittany

**Mush** · February 7th 2009, 04:40 PM

Originally Posted by williamb

consider the function

g(x) = [((x-8)^2)*((x-1)^3)] / [sqrt(x-5)]

i need to find the limits ..
x -> 8+
x-> 8-
x-> 5+
x-> 5-
x-> 1+
x-> 1-

i've found
x -> 8+ = infinity
x-> 8- = infinity
x-> 5+ = infinity
x-> 5- = DNE
x-> 1+ = DNE
x-> 1- = DNE

i thought that the last three didn't exsist because the square root was of a negative number when these were used, i know how to find limits but obviously i'm missing something here.
i was almost certain that sqrt(-) cannot be done for this,
please help me it's driving me mad

thanks in advance,
brittany

A limit exists for a function IFF $\displaystyle \lim_{x \to a^+} f(x) = \lim_{x \to a^-} f(x)$

So right away it doesn't make sense for the limit to exist for 3, but not 4.

So let's look at it:

$\displaystyle \lim_{x \to a} \frac{(x-8)^2(x-1)^3}{\sqrt(x-5)}$

$= \frac{ \bigg(\displaystyle \lim_{x \to a}(x-8)^2\bigg) \times \bigg(\displaystyle \lim_{x \to a} (x-1)^3 \bigg)}{ \displaystyle \lim_{x \to a} (\sqrt{x-5})}$

$= \frac{ \bigg(\displaystyle \lim_{x \to a}(x-8)\bigg)^2 \times \bigg(\displaystyle \lim_{x \to a} (x-1) \bigg)^3}{ \sqrt{\displaystyle \lim_{x \to a}(x-5)}}$

Now for $a = 8^+$ we get:

$= \frac{ (0)^2 \times (7)^3}{ \sqrt{(3)}} = 0$

$a = 8^-$ we get:

$= \frac{ (0)^2 \times (7)^3}{ \sqrt{(3)}} = 0$

$a = 5^+$ we get:

$= \frac{ (-3)^2 \times (4)^3}{ \sqrt{(0)}} = \pm \infty$

$a = 5^-$ is the one you had trouble with. Now you know that $5^-$ means it approaches 5 through negative values. But this doesn't always imply that we have a negative number. As x approaches 5 through negative values it must also pass through the interval $[0,5)$ , and these are non-negative. So we actually get:

$= \frac{ (-3)^2 \times (4)^3}{ \sqrt{(0)}} = \pm \infty$

And again, for $1^+$ and $1^-$ you are getting negative square root, so indeed, an illegal operation.

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