Thread: [SOLVED] L'Hopital's Rule &amp; One-Sided Limits.

1. [SOLVED] L'Hopital's Rule &amp; One-Sided Limits.

Help me please? I am getting confused by these one sided limits, and the text ihave just lightly touches this topic.

The question is : Use L'Hopital's Rule to find the limit:

$\lim_{ \mbox{y} \rightarrow 0^+} \frac{ \ln(6y^{2}+15y)}{ \ln y}$

and

$\lim_ {x \rightarrow \infty} \frac{ \ln(x+5)}{log_{7}x}$

What i didn't understand was the example in my text about this. It says that $\lim_{ \mbox{x} \rightarrow 0^+} \sqrt{x} \ln{x} = \lim_{ \mbox{x} \rightarrow 0^+} (-2 \sqrt{x}) = 0$ Why is this? I am a bit confused by the one-sided limits...

2. Originally Posted by Ife
Help me please? I am getting confused by these one sided limits, and the text ihave just lightly touches this topic.

The question is : Use L'Hopital's Rule to find the limit:

$\lim_{ \mbox{y} \rightarrow 0^+} \frac{ \ln(6y^{2}+15y)}{ \ln y}$

and

$\lim_ {x \rightarrow \infty} \frac{ \ln(x+5)}{log_{7}x}$

What i didn't understand was the example in my text about this. It says that $\lim_{ \mbox{x} \rightarrow 0^+} \sqrt{x} \ln{x} = \lim_{ \mbox{x} \rightarrow 0^+} (-2 \sqrt{x}) = 0$ Why is this? I am a bit confused by the one-sided limits...
Both $\ln(6y^2+15y)$ and $\ln{y}$ tend to $-\infty$ as $y\to0^+$. Therefore L'Hopital's Rule applies.

So taking the derivative of both the numerator and denominator, we have

$\lim_{ \mbox{y} \rightarrow 0^+} \frac{ \ln(6y^{2}+15y)}{ \ln y}=\lim_{y\to0^+}\frac{\frac{12y+15}{6y^2+15y}}{\f rac{1}{y}}=\lim_{y\to0^+}\frac{4y+5}{2y+5}$

Now, direct substitution yields

$L=1$.

3. Originally Posted by VonNemo19
Both $\ln(6y^2+15y)$ and $\ln{y}$ tend to $-\infty$ as $y\to0^+$. Therefore L'Hopital's Rule applies.

So taking the derivative of both the numerator and denominator, we have

$\lim_{ \mbox{y} \rightarrow 0^+} \frac{ \ln(6y^{2}+15y)}{ \ln y}=\lim_{y\to0^+}\frac{\frac{12y+15}{6y^2+15y}}{\f rac{1}{y}}=\lim_{y\to0^+}\frac{4y+5}{2y+5}$

Now, direct substitution yields

$L=1$.
Huh? how is the $\lim_{ \mbox{y} \rightarrow 0^+} \frac{ \ln(6y^{2}+15y)}{ \ln y}=\lim_{y\to0^+}\frac{\frac{12y+15}{6y^2+15y}}{\f rac{1}{y}}$? I dont understand the fraction in the numerator?

4. Originally Posted by Ife
Huh? how is the $\lim_{ \mbox{y} \rightarrow 0^+} \frac{ \ln(6y^{2}+15y)}{ \ln y}=\lim_{y\to0^+}\frac{\frac{12y+15}{6y^2+15y}}{\f rac{1}{y}}$? I applied L'Hopital's rule.

I dont understand the fraction in the numerator?
Because $\frac{d}{dx}[\ln{u}]=\frac{u'}{u}$ where $u$
is a differentiable function of $x$

5. Originally Posted by VonNemo19
Because $\frac{d}{dx}[\ln{u}]=\frac{u'}{u}$ where $u$
is a differentiable function of $x$
Thanks. I didn't know that much, i just knew about 1/x. But shouldn't the denominator remain as 16y + 15? how you simplify that to 2y+5? I am olny seeing that y cancels throughout when you invert and multiply and that's as simple as it'll get.. what am i missing?

6. Originally Posted by Ife
Thanks. I didn't know that much, i just knew about 1/x. But shouldn't the denominator remain as 12y + 15? how you simplify that to 2y+5? I am olny seeing that y cancels throughout when you invert and multiply and that's as simple as it'll get.. what am i missing?
I factored out and cancelled a 3.

7. Originally Posted by VonNemo19
I factored out and cancelled a 3.
S*@t! I had a 16 somehow on my paper here instead of a 6. Dunno where that 1 came from. Oops! lol

Thanks again (and sorry for the back and forth)

8. Originally Posted by Ife
S*@t! I had a 16 somehow on my paper here instead of a 6. Dunno where that 1 came from. Oops! lol

Thanks again (and sorry for the back and forth)
That's quite alright. That's what this forum is all about. If you don't understand something, then ask a question.

Are you all set then?

9. for the second limit just use L'Hospital's Rule. As for your question on the bottom, make sqrt(x)ln(x) into the equivalent fraction ln(x) / 1/sqrt(x). then use L'Hospital's Rule and you get 1/x / (-1/2)x^(-3/2), using the laws of exponents, when you divide two terms with like bases you subtract the powers so -1 - (-3/2) = 1/2 and 1 / (-1/2) becomes -2 so finally your expression now should looke like -2sqrt(x) and if you plug in 0, you get the limit as 0.

10. Originally Posted by VonNemo19
That's quite alright. That's what this forum is all about. If you don't understand something, then ask a question.

Are you all set then?
For this one, i guess. But honestly, the one-sided limit concept is very unclear to me, and i guess some of the limit rules associated with that and some trig functions and so. And the resourses i have are a bit limited, so I am not getting a clear pic at all. Then came this L'Hopital and Newton rules that I havent gotten much of either, which is just compounding what i don't know. I can't even understand the example in the book. See my 1st post on this thread. I quoted an example from my text. Can you explain that to me please?

11. Originally Posted by oblixps
for the second limit just use L'Hospital's Rule. As for your question on the bottom, make sqrt(x)ln(x) into the equivalent fraction ln(x) / 1/sqrt(x). then use L'Hospital's Rule and you get 1/x / (-1/2)x^(-3/2), using the laws of exponents, when you divide two terms with like bases you subtract the powers so -1 - (-3/2) = 1/2 and 1 / (-1/2) becomes -2 so finally your expression now should looke like -2sqrt(x) and if you plug in 0, you get the limit as 0.
I understand the calculation but why if the we are dealing with a one sided limit, isnt it that we are considering the values as they approach 0 from the right, but not the 0 itself? then how is the answer 0? Do you still just plug it in like you would a regular limit?

12. Originally Posted by Ife
For this one, i guess. But honestly, the one-sided limit concept is very unclear to me, and i guess some of the limit rules associated with that and some trig functions and so. And the resourses i have are a bit limited, so I am not getting a clear pic at all. Then came this L'Hopital and Newton rules that I havent gotten much of either, which is just compounding what i don't know. I can't even understand the example in the book. See my 1st post on this thread. I quoted an example from my text. Can you explain that to me please?
Understand that $\sqrt{x}$ is undefined for negative values, so $\lim_{x\to0}\sqrt{x}$ doesn't make any sense. But $\lim_{x\to0^+}\sqrt{x}$ makes perfect sense. We are still concerned with the behavior of $\sqrt{x}$ as $x\to0$, but we haven't made a s non-sensical statement.

Do you understand?

As for the example itself...

$\lim_{x\to0^+}\sqrt{x}\ln{x}$ gives the indeterminant form $0\cdot-\infty$, so rewriting we have

$\lim_{x\to0^+}\sqrt{x}\ln{x}=\lim_{x\to0^+}\frac{\ ln{x}}{\frac{1}{\sqrt{x}}}$ which now produces $-\frac{\infty}{\infty}$.We apply L'Hopital's rule and we have

$=\lim_{x\to0^+}\frac{\frac{1}{x}}{-\frac{1}{2\sqrt{x}^3}}=-\lim_{x\to0^+}{2}{\sqrt{x}}$

Which is what your text has.

13. Originally Posted by VonNemo19
Understand that $\sqrt{x}$ is undefined for negative values, so $\lim_{x\to0}\sqrt{x}$ doesn't make any sense. But $\lim_{x\to0^+}\sqrt{x}$ makes perfect sense. We are still concerned with the behavior of $\sqrt{x}$ as $x\to0$, but we haven't made a s non-sensical statement.

Do you understand?

As for the example itself...

$\lim_{x\to0^+}\sqrt{x}\ln{x}$ gives the indeterminant form $0\cdot-\infty$, so rewriting we have

$\lim_{x\to0^+}\sqrt{x}\ln{x}=\lim_{x\to0^+}\frac{\ ln{x}}{\frac{1}{\sqrt{x}}}$ which now produces $-\frac{\infty}{\infty}$.We apply L'Hopital's rule and we have

$=\lim_{x\to0^+}\frac{\frac{1}{x}}{-\frac{1}{2\sqrt{x}}}=-\lim_{x\to0^+}\frac{2}{\sqrt{x}}$

Which isn't quite what your text has.
Thanks. The text does have those steps, but my problem was just about the substitution of the 0 since it was a one sided limit. I wasn't sure that you could've simply plugged in the zero even though we are concerned with values above zero. But you explanation of that makes it clearer. Thanks. The other steps before i understood.

thanks.

14. Originally Posted by Ife
Thanks. The text does have those steps, but my problem was just about the substitution of the 0 since it was a one sided limit. I wasn't sure that you could've simply plugged in the zero even though we are concerned with values above zero. But you explanation of that makes it clearer. Thanks. The other steps before i understood.

thanks.
Ife, look at my post again. I've edited. I thought that I did it quick enough so that you wouldn't discover my mistake.

15. Originally Posted by Ife
I understand the calculation but why if the we are dealing with a one sided limit, isnt it that we are considering the values as they approach 0 from the right, but not the 0 itself? then how is the answer 0? Do you still just plug it in like you would a regular limit?
you're looking for what the function approaches as x approaches 0 from the right. if it will help you see, make a table of values and see what happens to the function when you plug in numbers approaching 0 from the right. you can also graph the function and observe the behavior as x approaches 0 from the right. when dealing with symbols, just plug in the number 0 since most of the time the limit as x approaches some some value is equal to the function evaluated at that value.