Find the limit of the product of ?(?) and ?(??) as ? → 0+. Clearly explain your steps.

[john458776]

I really dont know how I'm going to solve this problem I tried and I'm not sure if I'm correct. please help me I'm practicing for my upcoming test.

 

[Deleted member 4993]

View attachment 27211
I really dont know how I'm going to solve this problem I tried and I'm not sure if I'm correct. please help me I'm practicing for my upcoming test.

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem

 

[john458776]

I'm stuck here.

 

[skeeter]

from the shape of the graphs, looks like one can let [MATH]g(t) = \sin{t}[/MATH] and [MATH]k(t) = \ln{t}[/MATH]
[MATH]\lim_{t \to 0^+} \sin{t} \cdot \ln{t}[/MATH] can be expressed as [MATH]\lim_{t \to 0^+} \dfrac{\ln{t}}{\csc{t}}[/MATH]
should now be able to use L’Hopital …

 

[Dr.Peterson]

View attachment 27211
I really dont know how I'm going to solve this problem I tried and I'm not sure if I'm correct. please help me I'm practicing for my upcoming test.

The trouble is that \(0\cdot-\infty\) is indeterminate, so that the limit of the product depends on exactly how each factor approaches its own limit. It looks like \(\sin(x)\cdot\ln(x)\), but that's quite different from "is". If k deviates from \(\ln(x)\) in a significant way, the result may be entirely different.

So I don't think you can really find the limit; the best you can do is to guess it, as skeeter has suggested.

 

[john458776]

The limit will be zero if I use what skeeter has suggested.

 

[Deleted member 4993]

The limit will be zero if I use what skeeter has suggested.

Please show your work using "what skeeter has suggested ".

 

[skeeter]

zero, or indeterminate … discuss with your instructor as to how he/she sees it

 

[lex]

I really dont know how I'm going to solve this problem

You can't - it can't be answered.
E.g.
[MATH]g(x)=\sin \left(\frac{x}{10}\right)[/MATH], [MATH]\hspace2ex k(x)=-\frac{1}{\sqrt{x}}[/MATH], [MATH]\hspace2ex \lim \limits_{x \to 0^{+}} g(x)k(x)=0[/MATH]
[MATH]g(x)=\sin \left(\frac{x}{10}\right)[/MATH], [MATH]\hspace2ex k(x)=-\frac{1}{x}[/MATH], [MATH]\hspace2ex \lim \limits_{x \to 0^{+}} g(x)k(x)=-\frac{1}{10}[/MATH]
[MATH]g(x)=\sin \left(\frac{x}{10}\right)[/MATH], [MATH]\hspace2ex k(x)=-\frac{1}{e^{-\frac{1}{x}}}[/MATH], [MATH]\hspace2ex \lim \limits_{x \to 0^{+}} g(x)k(x)=-\infty[/MATH]

 

[lex]

(For the final example [MATH]k(x)=-e^{\frac{1}{x}}[/MATH] equivalently).

 

[skeeter]

(For the final example [MATH]k(x)=-e^{\frac{1}{x}}[/MATH] equivalently).

may want to add a positive constant to k(x) to get an x-intercept ... on all the examples

 

[lex]

may want to add a positive constant to k(x) to get an x-intercept ... on all the examples

Missed that. Thanks.

 

[lex]

(Repost of #9, incorporating skeeter's suggested amendment).

You can't - it can't be answered.
E.g.
[MATH]g(x)=\sin \left(\frac{x}{10}\right)[/MATH], [MATH]\hspace2ex k(x)=-\frac{1}{\sqrt{x}}+\frac{1}{2}[/MATH], [MATH]\hspace2ex \lim \limits_{x \to 0^{+}} g(x)k(x)=0[/MATH]
[MATH]g(x)=\sin \left(\frac{x}{10}\right)[/MATH], [MATH]\hspace2ex k(x)=-\frac{1}{x}+\frac{1}{2}[/MATH], [MATH]\hspace2ex \lim \limits_{x \to 0^{+}} g(x)k(x)=-\frac{1}{10}[/MATH]
[MATH]g(x)=\sin \left(\frac{x}{10}\right)[/MATH], [MATH]\hspace2ex k(x)=-e^{\frac{1}{x}}+\frac{3}{2}[/MATH], [MATH]\hspace2ex \lim \limits_{x \to 0^{+}} g(x)k(x)=-\infty[/MATH]

 

[nasi112]

(Repost of #9, incorporating skeeter's suggested amendment).

You can't - it can't be answered.
E.g.
[MATH]g(x)=\sin \left(\frac{x}{10}\right)[/MATH], [MATH]\hspace2ex k(x)=-\frac{1}{\sqrt{x}}+\frac{1}{2}[/MATH], [MATH]\hspace2ex \lim \limits_{x \to 0^{+}} g(x)k(x)=0[/MATH]View attachment 27233
[MATH]g(x)=\sin \left(\frac{x}{10}\right)[/MATH], [MATH]\hspace2ex k(x)=-\frac{1}{x}+\frac{1}{2}[/MATH], [MATH]\hspace2ex \lim \limits_{x \to 0^{+}} g(x)k(x)=-\frac{1}{10}[/MATH]View attachment 27234
[MATH]g(x)=\sin \left(\frac{x}{10}\right)[/MATH], [MATH]\hspace2ex k(x)=-e^{\frac{1}{x}}+\frac{3}{2}[/MATH], [MATH]\hspace2ex \lim \limits_{x \to 0^{+}} g(x)k(x)=-\infty[/MATH]View attachment 27235

That's beautiful, but I think that the main idea of this question is that to let the student to pick any two functions that are similar to the graph and proves his/her answer. It does not matter what is the answer.

 

[Dr.Peterson]

That's beautiful, but I think that the main idea of this question is that to let the student to pick any two functions that are similar to the graph and proves his/her answer. It does not matter what is the answer.

That's possible, but it's not an appropriate thing to do on a test, without explicitly stating it.

I dislike problems that penalize diligent students who want to be sure they are correct, and realize there is not one right answer (perhaps after a long struggle). If a problem says "find ___", it implies there is a specific ___ to be found.

Students who are used to guessing answers on tests, and assuming things they don't know, would have less trouble. That is not what should be taught in a math class.

 

[john458776]

That's possible, but it's not an appropriate thing to do on a test, without explicitly stating it.

I dislike problems that penalize diligent students who want to be sure they are correct, and realize there is not one right answer (perhaps after a long struggle). If a problem says "find ___", it implies there is a specific ___ to be found.

Students who are used to guessing answers on tests, and assuming things they don't know, would have less trouble. That is not what should be taught in a math class.

You are correct Dr.Peterson.

 

[skeeter]

maybe this was an open-ended question on a set of practice problems to foster class discussion …