A limit problem

**borophyll** · September 5th 2009, 10:47 PM

Find lim x->2 (f(x^2+5) - f(9))/(x-2), knowing that f'(x) = 1/x and f(2) = 9.

I am not sure how to solve this. I suppose f(x) = ln(x), but I don't think that helps me, since the textbook hasn't introduced natural logarithm yet, and from what has been covered so far, I am guessing that it can be solved some other way without this knowledge. Evaluating as is obviously generates a zero denominator.

And thats all I can say, I am completely stumped as to where to begin?

**mr fantastic** · September 5th 2009, 10:56 PM

Originally Posted by borophyll

Find lim x->2 (f(x^2+5) - f(9))/(x-2), knowing that f'(x) = 1/x and f(2) = 9.

I am not sure how to solve this. I suppose f(x) = ln(x), but I don't think that helps me, since the textbook hasn't introduced natural logarithm yet, and from what has been covered so far, I am guessing that it can be solved some other way without this knowledge. Evaluating as is obviously generates a zero denominator.

And thats all I can say, I am completely stumped as to where to begin?

There are many ways to do it but I'm not sure how it can be done at the level you claim to be at.

Even solving for f(x) (which is $\ln \left| \frac{x}{2}\right| + 9$ by the way) the limit will still have the indeterminant form 0/0.

The easiest way of doing it would be to use l'Hopital's Rule ....

**borophyll** · September 5th 2009, 11:08 PM

hmm, it is in the 'Challenging Problems' section of the chapter, but there has been no coverage of Lhopital's rule, and exponential and logarithmic functions aren't covered 'til CH 4. FYI, it is problem 2(a), challenging problems at end of Chapter 2 of Calculus 3/E (Robert A. Adams), if anyone has access to this text.

**Prove It** · September 5th 2009, 11:16 PM

Originally Posted by borophyll

Find lim x->2 (f(x^2+5) - f(9))/(x-2), knowing that f'(x) = 1/x and f(2) = 9.

I am not sure how to solve this. I suppose f(x) = ln(x), but I don't think that helps me, since the textbook hasn't introduced natural logarithm yet, and from what has been covered so far, I am guessing that it can be solved some other way without this knowledge. Evaluating as is obviously generates a zero denominator.

And thats all I can say, I am completely stumped as to where to begin?

Since this limit is indeterminate, i.e. $\frac{0}{0}$ , we can use L'Hospital's rule, which states that if a limit is indeterminate,

$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$ .

So in your case

$\lim_{x \to 2}\frac{f(x^2 + 5) - f(9)}{x - 2} = \lim_{x\to 2}\frac{\frac{d}{dx}[f(x^2 + 5) - f(9)]}{\frac{d}{dx}(x^2 + 5)}$ .

$\frac{d}{dx}[f(x^2 + 5) - f(9)] = f'(x^2 + 5) - f'(9) = \frac{1}{x^2 + 5} - \frac{1}{9}$ .

$\frac{d}{dx}(x^2 + 5) = 2x$ .

So $\lim_{x \to 2}\frac{f(x^2 + 5) - f(9)}{x - 2} = \lim_{x \to 2}\frac{\frac{1}{x^2 + 5} - \frac{1}{9}}{2x}$

$= \frac{\frac{1}{9} - \frac{1}{9}}{4}$

$= \frac{0}{4}$

$= 0$ .

**mr fantastic** · September 5th 2009, 11:24 PM

Originally Posted by Prove It

Since this limit is indeterminate, i.e. $\frac{0}{0}$ , we can use L'Hospital's rule, which states that if a limit is indeterminate,

$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$ .

So in your case

$\lim_{x \to 2}\frac{f(x^2 + 5) - f(9)}{x - 2} = \lim_{x\to 2}\frac{\frac{d}{dx}[f(x^2 + 5) - f(9)]}{\frac{d}{dx}(x^2 + 5)}$ .

$\frac{d}{dx}[f(x^2 + 5) - f(9)] = f'(x^2 + 5) - f'(9) = \frac{1}{x^2 + 5} - \frac{1}{9}$ . Mr F says: f(x^2 + 5) is a function of a function so the chain rule is needed to differentiate with respect to x: df/dx = 2x f'(x^2 + 5). f(9) is a constant so its derivative is zero. Therefore the derivative of the numerator is 2x f'(x^2 + 5) and the limit of this as x --> 2 is 4 f'(9) = 4/9.

$\frac{d}{dx}(x^2 + 5) = 2x$ . Mr F says: The denominator is (x - 2) so the derivative with respect to x is 1.

So $\lim_{x \to 2}\frac{f(x^2 + 5) - f(9)}{x - 2} = \lim_{x \to 2}\frac{\frac{1}{x^2 + 5} - \frac{1}{9}}{2x}$

$= \frac{\frac{1}{9} - \frac{1}{9}}{4}$

$= \frac{0}{4}$

$= 0$ .

A couple of corrections (in red). I get 4/9 as the final answer.

**Chris L T521** · September 5th 2009, 11:30 PM

Originally Posted by borophyll

Find lim x->2 (f(x^2+5) - f(9))/(x-2), knowing that f'(x) = 1/x and f(2) = 9.

I am not sure how to solve this. I suppose f(x) = ln(x), but I don't think that helps me, since the textbook hasn't introduced natural logarithm yet, and from what has been covered so far, I am guessing that it can be solved some other way without this knowledge. Evaluating as is obviously generates a zero denominator.

And thats all I can say, I am completely stumped as to where to begin?

Let $u=x^2+5$ . Thus as $x\to2,\,u\to9$ .

So we have $\lim_{u\to9}\frac{f(u)-f(9)}{\sqrt{u-5}-2}=\lim_{u\to9}\frac{f(u)-f(9)}{(u-5)-4}\cdot\left(\sqrt{u-5}+2\right)=\lim_{u\to9}\frac{f(u)-f(9)}{u-9}\cdot\left(\sqrt{u-5}+2\right)$ $=f^{\prime}(9)\cdot\lim_{u\to9}\left(\sqrt{u-5}+2\right)=f^{\prime}(9)\cdot(4)=\frac{4}{9}$

Does this look right?

**borophyll** · September 5th 2009, 11:38 PM

OK, thanks for that. But it still begs the question: why would the author have a problem that is impossible to solve with the expected prerequisite knowledge? The entire textbook does not have a single mention of L'hopitals rule at all.

**borophyll** · September 5th 2009, 11:52 PM

Originally Posted by Chris L T521

Let $u=x^2+5$ . Thus as $x\to2,\,u\to9$ .

So we have $\lim_{u\to9}\frac{f(u)-f(9)}{\sqrt{u-5}-2}=\lim_{u\to9}\frac{f(u)-f(9)}{(u-5)-4}\cdot\left(\sqrt{u-5}+2\right)=\lim_{u\to9}\frac{f(u)-f(9)}{u-9}\cdot\left(\sqrt{u-5}+2\right)$ $=f^{\prime}(9)\cdot\lim_{u\to9}\left(\sqrt{u-5}+2\right)=f^{\prime}(9)\cdot(4)=\frac{4}{9}$

Does this look right?

I don't understand how you converted $\frac{f(u)-f(9)}{u-9}$ to $f^{\prime}(9)$ ?

**mr fantastic** · September 6th 2009, 01:37 AM

Originally Posted by borophyll

I don't understand how you converted $\frac{f(u)-f(9)}{u-9}$ to $f^{\prime}(9)$ ?

One of the definitions of the derivative of f(x) at x = a is $\lim_{x \rightarrow a} \frac{f(x) - f(a)}{x - a}$ .

Originally Posted by borophyll

OK, thanks for that. But it still begs the question: why would the author have a problem that is impossible to solve with the expected prerequisite knowledge? The entire textbook does not have a single mention of L'hopitals rule at all.

It's clear from post #6 that l'Hopital's Rule is not required.

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