I have a problem evaluating the following limit:
\(\displaystyle $\lim_{x\rightarrow-4} \frac{1/4 + 1/x}{x+4}$\)
Now, I have tried multiplying the expression by 1/1 equivalents of many types: \(\displaystyle $\frac{x-4}{x-4}$, $\frac{1/x - 1/4}{1/x - 1/4}$\), and many combinations thereof.
But every time the expression seems to reduce to 0/0 !!
Now, I understand that I might use L'Hospital's rule to evaluate the limit (although, in the book I'm working from, L'Hospital's Rule is not introduced until a later chapter).
The book, my TI-89 graphing calculator, and Wolfram Alpha all give the limit as -1/16.
heres the Wolfram Alpha link: http://www.wolframalpha.com/input/?i=lim((1/4%2B1/x)/(4%2Bx))+as+x-%3E+-4
L'Hospital's rule says that if we have a 0/0 expression, we can find the quotient by dividing the derivative of the numerator by the derivative of the denominator.
So if you have F(x) / G(x) and the ratio is 0/0, you can substitute F'(x) / G'(x) . That much I understand. Where I'm messing up seems to be the definition of G(x).
Wolfram Alpha does the rule like this:
it calls the numerator \(\displaystyle F(x) = 1/x + 1/4\)
So far so good.
But it then it defines \(\displaystyle G(x) = \frac{1}{x+4}\) or \(\displaystyle G(x) = (x+4)^-1\). Then it finds the derivative \(\displaystyle G'(x) = \frac{1}{x^2}\) or \(\displaystyle x^-2\)
That's what I'm not getting. Doesn't L'Hospital's Rule say that you should treat the denominator of the indeterminate form as a function G(x) by itself, and not underneath a numerator? In other words, my understanding is that the we should take
\(\displaystyle G(x) = x+4\) so \(\displaystyle G'(x) = 1\) ,
not \(\displaystyle G(x) = \frac{1}{x+4} ---> G'(x) = \frac{1}{x^2}\)
As Wolfram Alpha does it, it gets the same answer as the book and my TI89: -1/16.
The answer looks correct: \(\displaystyle $\lim_{x\rightarrow-4} \frac{1/x+1/4}{x+4} = \frac{F(x)}{G(x)} = \frac{F'(x)}{G'(x)} = -1/x^2 = -1/16 $\)
Where am I going wrong? Why is \(\displaystyle G(x) = \frac{1}{x+4}\) and not just \(\displaystyle G(x) = x+4\) ???
\(\displaystyle $\lim_{x\rightarrow-4} \frac{1/4 + 1/x}{x+4}$\)
Now, I have tried multiplying the expression by 1/1 equivalents of many types: \(\displaystyle $\frac{x-4}{x-4}$, $\frac{1/x - 1/4}{1/x - 1/4}$\), and many combinations thereof.
But every time the expression seems to reduce to 0/0 !!
Now, I understand that I might use L'Hospital's rule to evaluate the limit (although, in the book I'm working from, L'Hospital's Rule is not introduced until a later chapter).
The book, my TI-89 graphing calculator, and Wolfram Alpha all give the limit as -1/16.
heres the Wolfram Alpha link: http://www.wolframalpha.com/input/?i=lim((1/4%2B1/x)/(4%2Bx))+as+x-%3E+-4
L'Hospital's rule says that if we have a 0/0 expression, we can find the quotient by dividing the derivative of the numerator by the derivative of the denominator.
So if you have F(x) / G(x) and the ratio is 0/0, you can substitute F'(x) / G'(x) . That much I understand. Where I'm messing up seems to be the definition of G(x).
Wolfram Alpha does the rule like this:
it calls the numerator \(\displaystyle F(x) = 1/x + 1/4\)
So far so good.
But it then it defines \(\displaystyle G(x) = \frac{1}{x+4}\) or \(\displaystyle G(x) = (x+4)^-1\). Then it finds the derivative \(\displaystyle G'(x) = \frac{1}{x^2}\) or \(\displaystyle x^-2\)
That's what I'm not getting. Doesn't L'Hospital's Rule say that you should treat the denominator of the indeterminate form as a function G(x) by itself, and not underneath a numerator? In other words, my understanding is that the we should take
\(\displaystyle G(x) = x+4\) so \(\displaystyle G'(x) = 1\) ,
not \(\displaystyle G(x) = \frac{1}{x+4} ---> G'(x) = \frac{1}{x^2}\)
As Wolfram Alpha does it, it gets the same answer as the book and my TI89: -1/16.
The answer looks correct: \(\displaystyle $\lim_{x\rightarrow-4} \frac{1/x+1/4}{x+4} = \frac{F(x)}{G(x)} = \frac{F'(x)}{G'(x)} = -1/x^2 = -1/16 $\)
Where am I going wrong? Why is \(\displaystyle G(x) = \frac{1}{x+4}\) and not just \(\displaystyle G(x) = x+4\) ???