Substitution in One-Sided Limits

[Mampac]

Good day,

I'm a CS freshman who has Calculus I as a core requirement and limits confuse me a bit. My professor doesn't have office hours on the week-ends so I'd like to ask you something about one-sided limits.

I know that if x tends to n, then I have the full right to replace x with n, right? (unless some restrictions like roots and logarithms are an issue) (correct me if i'm mistaken)

Is the same rule applicable when x tends to n from one side? for example, if x --> 0+, can i replace x with 0? or should i replace it with -inf/+inf?

some problems occurred when i was computing a limit and had to use L'Hopital's rule (because my professor did the same during a problem solving session).

i rewrote sqrt(x)ln(tanx) as sqrt(x) as x^(1/2) / (ln(tanx))^(-1) in order to create a fraction.

now I need to use L'Hopital's, right? but if i replace x with 0 then i will get 0/undefined or 0/-inf. Therefore, I can't use it.

but my professor and TA's somehow did... someone can clear it up for me?

 

[Steven G]

If x --> 0+ why would you ever consider replacing x with infinity or negative infinity. I would think that replacing x with .000001 would be more accurate than replacing x with +/- infinity. Is it that you think that sometimes 0 is infinity or that 0 is -infinity?

To answer your question, you should replace x with 0 and see if you get a nice answer.


Can you show us what your work using L'Hopital's rule?

 

[Dr.Peterson]

I know that if x tends to n, then I have the full right to replace x with n, right? (unless some restrictions like roots and logarithms are an issue) (correct me if i'm mistaken)

Well, you can try that substitution, but if it gives you a value, you have to consider issues like continuity, which may nullify your "right"!

Is the same rule applicable when x tends to n from one side? for example, if x --> 0+, can i replace x with 0? or should i replace it with -inf/+inf?

If you know the function is continuous to the right and including 0, you can try putting in 0; no other substitution would be appropriate.

i rewrote sqrt(x)ln(tanx) as sqrt(x) as x^(1/2) / (ln(tanx))^(-1) in order to create a fraction.

now I need to use L'Hopital's, right? but if i replace x with 0 then i will get 0/undefined or 0/-inf. Therefore, I can't use it.

but my professor and TA's somehow did... someone can clear it up for me?

This is a different question than direct substitution. With L'Hopital, you plug in numbers (or, really, take limits of parts of the expression) to check the form of the limit, which in this case, [MATH]\sqrt{x}\ln(\tan x)[/MATH], looks like [MATH]0\ln(0) = (0)(-\infty)[/MATH], and then you have to do something like what you did to put it in the form [MATH]0/0[/MATH] or [MATH]\infty/\infty[/MATH]. I'd say that the form you chose looks like [MATH]0/0[/MATH], since [MATH](\ln(\tan x))^{-1}[/MATH] approaches 0. It's not what you get on actual substitution, but the limits of the numerator and denominator.

So, yes, proceed with L'Hopital. And maybe if that form gets really ugly (I haven't tried it yet), try something else like [MATH]\frac{\ln(\tan x)}{x^{-1/2}}[/MATH].

 

[Mampac]

...[MATH](\ln(\tan x))^{-1}[/MATH] approaches 0...

Could you explain why it tends to 0 if tan0 = 0 and ln(0) is undefined?

I tried both ways of writing and one of them, the original one i mentioned, when I raised ln(tanx) to the power -1, got very dirty because I couldn't get rid of 0 in the denominator since x to the power of something seems impossible to eradicate:



This one you suggested, though, looks cleaner... Do you think it's correct?

 

[Deleted member 4993]

Could you explain why it tends to 0 if tan0 = 0 and ln(0) is undefined?

I tried both ways of writing and one of them, the original one i mentioned, when I raised ln(tanx) to the power -1, got very dirty because I couldn't get rid of 0 in the denominator since x to the power of something seems impossible to eradicate:

View attachment 23135

This one you suggested, though, looks cleaner... Do you think it's correct?
View attachment 23136

You check your "answer" by using Wolframalpha.com

 

[Mampac]

You check your "answer" by using Wolframalpha.com

haha, y'know what, I tried on a different site (Symbolab)! and it spat out "steps are currently not available for this problem"!
thank you so much for help!
btw i just found out i've left another thread about a problem from SAT prep a year ago when i was applying to college... (now i'm again seeking help but in college haha) i never opened it again tho, but there were lots of answers and 2 people again commented on this one too! (including you) it's so cute damn.

 

[Dr.Peterson]

This one you suggested, though, looks cleaner... Do you think it's correct?
View attachment 23136

There's an error in your handling of the 2; be careful! By luck it doesn't affect the answer.

Also, I would have rewritten [MATH]\tan(x)\cos^2(x)[/MATH] as [MATH]\sin(x)\cos(x)[/MATH], and then as [MATH]\frac{1}{2}\sin(2x)[/MATH] to simplify the work a bit.

A graph shows that the answer is correct:

 

[Steven G]

In your 1st attempt you did not compute the derivative of (ln(tanx) )^-1 correctly.

 

[Steven G]

In your 2nd attempt you have a minor notation error. Once you replace x with 0 you should not write limit anymore.

 

[Mampac]

There's an error in your handling of the 2; be careful! By luck it doesn't affect the answer.

Also, I would have rewritten [MATH]\tan(x)\cos^2(x)[/MATH] as [MATH]\sin(x)\cos(x)[/MATH], and then as [MATH]\frac{1}{2}\sin(2x)[/MATH] to simplify the work a bit.

A graph shows that the answer is correct:
View attachment 23143

handling of which one of the 2s? haha
thank you all so much! i think i got the idea of limits now

 

[Dr.Peterson]

handling of which one of the 2s? haha
thank you all so much! i think i got the idea of limits now

You went from a -1/2 on the bottom to a 2 on the bottom; that 2 should have moved to the top, as 1/(1/2) = 2/1, not 1/2.