Evaluating limits with only variables

[greystreet34]

Hi, I'm new here and this is my first post but I'm having trouble with one of my homework problems. I have the function:

g(x) = (a(x+b)(x-c)(x+d)) / (e(x-f)(x-b)(x-c)), where a, b, c, d, e, and f are all unique, positive real numbers.

a) At what values of x=h does the limit g(x) = -∞ or +∞ as x->h?

b) Evaluate the limit of g(x) as x->c.

c) Does the limit g(x) as x->∞ exist?

I'm completely stuck and unsure where to start... I'm tempted to cross out the (x-c) from both the numerator and denominator but that seems like a trick.

Thank you for any advice!

 

[DrPhil]

Hi, I'm new here and this is my first post but I'm having trouble with one of my homework problems. I have the function:

g(x) = (a(x+b)(x-c)(x+d)) / (e(x-f)(x-b)(x-c)), where a, b, c, d, e, and f are all unique, positive real numbers.

a) At what values of x=h does the limit g(x) = -∞ or +∞ as x->h?

b) Evaluate the limit of g(x) as x->c.

c) Does the limit g(x) as x->∞ exist?

I'm completely stuck and unsure where to start... I'm tempted to cross out the (x-c) from both the numerator and denominator but that seems like a trick.

Thank you for any advice!

Definitely DO cancel any factors that are equal in numerator and denominator! Then

\(\displaystyle g(x) = \dfrac{a (x + b)(x + d)}{e (x - f)(x - b)}\)

a) wherever the denominator -->0, then then g(x) increases without limit

b) Once the factors (x-c) have been cancelled, g(c) is perfectly well defined.

c) Do you know l'Hospital's Rule? If not, can you see what happens when x>>{b,d, f} ?

 

[greystreet34]

So part a is for the values where x=f or x=b? For part b, what do you mean by g(c) is perfectly well defined?

 

[JeffM]

So part a is for the values where x=f or x=b? The function is defined at neither f nor b. For part b, what do you mean by g(c) is perfectly well defined?

If we are being strict with our definitions

\(\displaystyle g(x) = \dfrac{a(x + b)(x - c)(x + d)}{e(x - f)(x - b)(x - c)}\)

does not exist (is not defined) if x = f, x = b, or x = c.

However,

\(\displaystyle x \ne b,\ x \ne c,\ and\ x \ne f \implies g(x) = \dfrac{a(x + b)(x - c)(x + d)}{e(x - f)(x - b)(x - c)} = \dfrac{a(x + b)(x + d)}{e(x - f)(x - b)} = h(x).\)

Now when we are taking the limit as the argument of a function is approaching some number, we exclude from consideration the function of that number.

\(\displaystyle \displaystyle \lim_{x \rightarrow c}g(x) = \lim_{x \rightarrow c}h(x) = \lim_{x \rightarrow c}\dfrac{a(x + b)(x + d)}{e(x - f)(x - b)}.\)

The latter limit is perfectly well defined and so therefore is the former.

Remember this: even though a function may not be defined at a point, its limit may be defined because, technically, the limit is never AT the point, just very close to the point.

 

[greystreet34]

So you're saying the answer of part b is formatted as

lim g(x) = (a(x+b)(x+d)) / (e(x−f)(x−b))
x→c​

Also, I was thinking about part c. Doesn't lim g(x) as x->∞ look for values where x is really large? If I expand the numerator and denominator I get:

(a(x^2+bx+dx+bd)) / (e(x^2-bx-fx+bf)), which for large x leaves me with ax^2 / ex^2? Does that mean the limit exists and is a/e?

 

[DrPhil]

So you're saying the answer of part b is formatted as

lim g(x) = (a(x+b)(x+d)) / (e(x−f)(x−b))
x→c​

Also, I was thinking about part c. Doesn't lim g(x) as x->∞ look for values where x is really large? If I expand the numerator and denominator I get:

(a(x^2+bx+dx+bd)) / (e(x^2-bx-fx+bf)), which for large x leaves me with ax^2 / ex^2? Does that mean the limit exists and is a/e?

Since you can show the limit as x-->c is x=c in every factor, you can substitute x=c.

Yes, the limit as x--> oo exists, and is equal to a/e.

 

[JeffM]

So you're saying the answer of part b is formatted as

lim g(x) = (a(x+b)(x+d)) / (e(x−f)(x−b))
x→c​

Also, I was thinking about part c. Doesn't lim g(x) as x->∞ look for values where x is really large? If I expand the numerator and denominator I get:

(a(x^2+bx+dx+bd)) / (e(x^2-bx-fx+bf)), which for large x leaves me with ax^2 / ex^2? Does that mean the limit exists and is a/e?

It is not a matter of "formatting." It is a matter of understanding the subtle conceptual difference between

\(\displaystyle f(u)\ and\ \displaystyle \lim_{x \rightarrow u}f(x).\)

\(\displaystyle If\ and\ only\ if\ f(u) \in \mathbb R,\ \displaystyle \lim_{x \rightarrow u} \in \mathbb R,\ and\ f(u) = lim_{x \rightarrow u}f(x),\ then f(x)\ is\ continuous\ at\ u.\)

Notice that the definition implies that f(u) may not exist but that the limit of f(x) as x approaches u may exist. Or f(u) may exist, but the limit may not. Or the function may exist but not equal the limit. The limit of the function and the function are equal at a point only in some cases, namely in the case of a function that is continuous at that point.

So \(\displaystyle g(x) = \dfrac{a(x + b)(x - c)(x + d)}{e(x - b)(x - c)(x - f)}\ does\ not\ exist\ at\ x = b,\ c,\ or\ f.\)

And \(\displaystyle h(x) = \dfrac{a(x + b)(x + d)}{e(x - b)(x - f)}\ does\ not\ exist\ at\ x = b\ or\ f,\ but\ does\ exist\ at\ x = c.\)

And \(\displaystyle g(x) = h(x)\ unless\ x = b,\ c,\ or\ f.\)

Because, technically, we compute the limit of f(x) as x approaches c at values of x close to but not equal to c, we can say

\(\displaystyle \displaystyle \lim_{x \rightarrow c}g(x) = \lim_{x \rightarrow c}h(x) = \dfrac{a(c + b)(c + d)}{e(c - b)(c - f)}.\)

It is not formatting; it is an equality because the two functions are equal (not identical) except at three points.

You are analyzing c correctly.