- #1
honestrosewater
Gold Member
- 2,142
- 6
Problem
Find the limit (if it exists). If it does not, explain why.
[tex]
\lim_{x \rightarrow -3^{-}} \frac{x}{\sqrt{x^{2}-9}}
[/tex]
(i.e. the limit from the left.)
Definitions & theorems
vertical asymptote definition: if f(x) approaches infinity (or negative infinity) as x approaches c from the right or the left, then the line x = c is a vertical asymptote of the graph of f.
vertical asymptote theorem: let f and g be continuous on an open interval containing c. If f(c) != 0, g(c) = 0, and there exists an open interval containing c such that g(x) != 0 for all x != c in the interval, then the graph of the function given by h(x) = f(x)/g(x) has a vertical asymptote at x = c.
infinite limit definition: let f be a function that is defined at every real number in some open interval containing c (except possibly at c itself). The statement
[tex]
\lim_{x \rightarrow c} f(x) = \infty
[/tex]
means that for each M > 0 there exists a d > 0 such that f(x) > M whenever 0 < |x-c| < d. [Limit for negative infinity is defined similarly.]
To define the infinite limit from the left, replace 0< |x-c| < d by c-d<x<c. [Limit from right is defined similarly.]
My attempt
(I am trying to do things as the book says. Personally, I hate this book.) A graph and my intuition say that there should be a vertical asymptote at -3, but I do not see how to prove it using the definitions and theorems in the book.
h(x) = x/sqrt(x^2 - 9) ; x^2 - 9 >= 0, sqrt(x^2 - 9) != 0
domain(h) = (-infinity, -3) U (3, infinity)
h(x) = f(x)/g(x) --> f(x) = x and g(x) = sqrt(x^2 - 9)
domain(f) = R, domain(g) = (-infinity, -3] U [3, infinity)
f(-3) != 0, g(-3) = 0
However, g is not defined on any open interval containing -3, so I do not know how the book wants me to proceed.
My answer would be that the limit does not exist because there is a vertical asymptote at x = -3 if I could figure out how to prove it. I will need to do so for other problems, so even if I can get away without it here, I need to know eventually.
Find the limit (if it exists). If it does not, explain why.
[tex]
\lim_{x \rightarrow -3^{-}} \frac{x}{\sqrt{x^{2}-9}}
[/tex]
(i.e. the limit from the left.)
Definitions & theorems
vertical asymptote definition: if f(x) approaches infinity (or negative infinity) as x approaches c from the right or the left, then the line x = c is a vertical asymptote of the graph of f.
vertical asymptote theorem: let f and g be continuous on an open interval containing c. If f(c) != 0, g(c) = 0, and there exists an open interval containing c such that g(x) != 0 for all x != c in the interval, then the graph of the function given by h(x) = f(x)/g(x) has a vertical asymptote at x = c.
infinite limit definition: let f be a function that is defined at every real number in some open interval containing c (except possibly at c itself). The statement
[tex]
\lim_{x \rightarrow c} f(x) = \infty
[/tex]
means that for each M > 0 there exists a d > 0 such that f(x) > M whenever 0 < |x-c| < d. [Limit for negative infinity is defined similarly.]
To define the infinite limit from the left, replace 0< |x-c| < d by c-d<x<c. [Limit from right is defined similarly.]
My attempt
(I am trying to do things as the book says. Personally, I hate this book.) A graph and my intuition say that there should be a vertical asymptote at -3, but I do not see how to prove it using the definitions and theorems in the book.
h(x) = x/sqrt(x^2 - 9) ; x^2 - 9 >= 0, sqrt(x^2 - 9) != 0
domain(h) = (-infinity, -3) U (3, infinity)
h(x) = f(x)/g(x) --> f(x) = x and g(x) = sqrt(x^2 - 9)
domain(f) = R, domain(g) = (-infinity, -3] U [3, infinity)
f(-3) != 0, g(-3) = 0
However, g is not defined on any open interval containing -3, so I do not know how the book wants me to proceed.
My answer would be that the limit does not exist because there is a vertical asymptote at x = -3 if I could figure out how to prove it. I will need to do so for other problems, so even if I can get away without it here, I need to know eventually.