# L'Hopital/limits question

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• January 8th 2011, 10:57 AM
durrrrrrrr
L'Hopital/limits question
Here's the question:
lim(x-->infinity) $x/((1+x^2)^(1/2))$

One approach (that Wolfram Alpha used):
=square root of lim $x^2/(1+x^2) = 1$
(using L'hopital)

Another approach:
Using L'hopital: $=((1+x^2)^(1/2))/x$
which is our original question upside down.
Does this prove the limit is 1.

ie. if limit a/b = limit b/a does this mean the limit must equal 1?

Thanks
• January 8th 2011, 11:07 AM
Plato
$\[\dfrac{x}{{\sqrt {1 + x^2 } }} = \dfrac{1}{{\sqrt {\frac{1}{{x^2 }} + 1} }}$
• January 8th 2011, 11:07 AM
Also sprach Zarathustra
No. Try to do the same to f(x)=x/e^x (for your first question!)
• January 8th 2011, 11:09 AM
snowtea
Quote:

Originally Posted by durrrrrrrr
ie. if limit a/b = limit b/a does this mean the limit must equal 1?

If both limits exist

limit a/b = 1/(limit b/a)

So limit a/b could also be -1.
• January 8th 2011, 11:20 AM
FernandoRevilla
Quote:

Originally Posted by Plato
$\[\dfrac{x}{{\sqrt {1 + x^2 } }} = \dfrac{1}{{\sqrt {\frac{1}{{x^2 }} + 1} }}$

It should be:

$f(x)=\dfrac{x}{{\sqrt {1 + x^2 } }} =\begin{Bmatrix} \dfrac{1}{{\sqrt {\frac{1}{{x^2 }} + 1} }} & \mbox{ if }& x> 0\\-\dfrac{1}{{\sqrt {\frac{1}{{x^2 }} + 1} }} & \mbox{if}& x<0\end{matrix}$

So,

$\displaystyle\lim_{x \to{+}\infty}{f(x)}=1,\quad \displaystyle\lim_{x \to{-}\infty}{f(x)}=-1$

and does not exist

$\displaystyle\lim_{x \to{}\infty}{f(x)}$

Fernando Revilla
• January 8th 2011, 11:46 AM
dwsmith
$\displaystyle\lim_{x\to\infty}\frac{x}{\sqrt{x^2+1 }}\Rightarrow\lim_{x\to\infty}\frac{x}{\sqrt{x^2}} \Rightarrow\lim_{x\to\infty}\frac{x}{\pm x}$

$\displaystyle\lim_{x\to\infty}\frac{x}{x}=1 \ \mbox{and} \ \lim_{x\to\infty}\frac{x}{-x}=-1$
• January 8th 2011, 11:48 AM
Plato
Because the OP is $\displaystyle \lim _{x \to \infty } \frac{x}{{\sqrt {1 + x^2 } }}$, we can assume that $x>0$
• January 8th 2011, 11:59 AM
FernandoRevilla
Quote:

Originally Posted by Plato
Because the OP is $\displaystyle \lim _{x \to \infty } \frac{x}{{\sqrt {1 + x^2 } }}$, we can assume that $x>0$

No, you can't assume that. The formula that defines $f$ has sense on $\mathbb{R}$ .When a function is given by a formula an the domain is not specified it is an universal convenious to apply the rule of the maximum domain i.e, $D(f)=\mathbb{R}$, and topologically $\infty\neq +\infty$ .

Fernando Revilla
• January 8th 2011, 12:02 PM
dwsmith
Quote:

Originally Posted by Plato
Because the OP is $\displaystyle \lim _{x \to \infty } \frac{x}{{\sqrt {1 + x^2 } }}$, we can assume that $x>0$

If we want to assume x>0, then shouldn't the limit be written as $\displaystyle\lim_{x\to\infty^+}\frac{x}{\sqrt{x^2 +1}}\mbox{?}$
• January 8th 2011, 12:04 PM
FernandoRevilla
Quote:

Originally Posted by dwsmith
If we want to assume x>0, then shouldn't the limit be written as $\displaystyle\lim_{x\to\infty^+}\frac{x}{\sqrt{x^2 +1}}\mbox{?}$

Right.

Fernando Revilla
• January 8th 2011, 12:05 PM
dwsmith
Quote:

Originally Posted by FernandoRevilla

Hence, the limit DNE since the (+) is infinity and (-) is negative infinity.
• January 8th 2011, 12:08 PM
FernandoRevilla
Quote:

Originally Posted by dwsmith
Hence, the limit DNE since the (+) is infinity and (-) is negative infinity.

Right.

Fernando Revilla
• January 8th 2011, 12:11 PM
Plato
Quote:

Originally Posted by FernandoRevilla
No, you can't assume that. The function $f$ is defined on $\mathbb{R}$ .When a function is given by a formula an the domain is not specified it is an universal convenious to apply the rule of the maximum domain i.e, $D(f)=\mathbb{R}$, and topologicaly $\infty\neq +\infty$ .

With all due respect, I think you have no idea about how this material is currently being taught. What you have written above is not consistent with any standard calculus text book in use today. I have taught calculus classes since 1964.
• January 8th 2011, 12:14 PM
Archie Meade
Does not the notation

$\displaystyle\lim_{x \to a^{+}}$

denote the limit as "x" approaches "a" from the right?

In which case

$\displaystyle\lim_{x \to \infty^{+}}$

requires x<0 and approaching $-\infty$
• January 8th 2011, 12:16 PM
dwsmith
Quote:

Originally Posted by Plato
With all due respect, I think you have no idea about how this material is currently being taught. What you have written above is not consistent with any standard calculus text book in use today. I have taught calculus classes since 1964.

Out of curiousity you are saying when infinity is printed in a limit in a Calculus text book, the book is implying the positive values only?

Also, my Calc text book isn't written in such a manner. If there isn't a + or -, the it is the overall limit. My book is very good about specifying direction.
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