Limit of a sine exercise

[wolly]

[math]\lim_{x \to -1} \frac{\sqrt[3]{x}+1}{sin\pi(x+1)}[/math]
Hi I want to solve this limit.What are the steps?
I know that [math]\lim_{x \to 0} \frac{sin(x)}{x} =1[/math]Also how can square roots and cube roots be negative when their condition îs x=>0?

 

[mario99]

[math]\lim_{x \to -1} \frac{\sqrt[3]{x}+1}{sin\pi(x+1)}[/math]
Hi I want to solve this limit.What are the steps?
I know that [math]\lim_{x \to 0} \frac{sin(x)}{x} =1[/math]Also how can square roots and cube roots be negative when their condition îs x=>0?

What have you done so far?

 

[wolly]

I replaced x with t-1.I used a limit calculator from the internet but I don't know If this îs how You solve it.

 

[wolly]

[math]x=t-1[/math]How did this answer appeared?

 

[Dr.Peterson]

I replaced x with t-1.I used a limit calculator from the internet but I don't know If this îs how You solve it.

Don't use anything but your own mind.

But you might try substituting [imath]u[/imath] for [imath]\sqrt[3]{x}[/imath], and be prepared to factor a sum of cubes ...

(I haven't carried this out, so there are no guarantees.)

 

[mario99]

I replaced x with t-1.I used a limit calculator from the internet but I don't know If this îs how You solve it.

I did not try [imath]t - 1[/imath]. Watch how I did it.

You said that [imath]\displaystyle \lim_{x\rightarrow 0} \frac{\sin x}{x} = 1[/imath].

Did you know also [imath]\displaystyle \lim_{x\rightarrow 0} \frac{x}{\sin x} = 1?[/imath]

Therefore,

[imath]\displaystyle \lim_{x\rightarrow -1} \frac{\sqrt[3]{x} + 1}{\sin(\pi[x + 1])} = \lim_{x\rightarrow -1} \frac{\pi[x + 1]}{\sin(\pi[x + 1])} \times \frac{\sqrt[3]{x} + 1}{\pi[x + 1]} = \frac{1}{\pi}\lim_{x\rightarrow -1} \frac{\sqrt[3]{x} + 1}{x + 1}[/imath].

Now it is a perfect moment to use L'hopital rule

[imath]\displaystyle \frac{1}{\pi}\lim_{x\rightarrow -1} \frac{\sqrt[3]{x} + 1}{x + 1} = \ ?[/imath]

If you did not like my method, try to use professor Dave's hint.

 

[wolly]

Don't use anything but your own mind.

But you might try substituting [imath]u[/imath] for [imath]\sqrt[3]{x}[/imath], and be prepared to factor a sum of cubes ...

(I haven't carried this out, so there are no guarantees.)

Do You replace [math]\sqrt[3]{x}=t[/math] or [math]\sqrt[3]{x}+1=t[/math]?
I'm confused!

 

[Dr.Peterson]

Do You replace [math]\sqrt[3]{x}=t[/math] or [math]\sqrt[3]{x}+1=t[/math]?
I'm confused!

My suggestion was

you might try substituting [imath]u[/imath] for [imath]\sqrt[3]{x}[/imath], and be prepared to factor a sum of cubes ...

so that answers your question; but then, if one of those didn't work, you could try the other. Don't be confused; just try!

If you make my substitution, what is x equal to? What is it if you do yours (the second)? I have no idea whether the latter will work, so I can't tell you not to try it.

The key to solving challenging problems is to try whatever occurs to you. Don't wait to know what is the "right" method before doing anything. And often, when you have taken one step, you can see a next step more clearly.

 

[lookagain]

[math]\lim_{x \to -1} \frac{\sqrt[3]{x}+1}{sin\pi(x+1)}[/math]

wolly, if you are sending this problem to any other mathematics message boards for help, please include
needed grouping symbols in the denominator.

\(\displaystyle \lim_{x \to -1} \frac{ \sqrt[3]{x} + 1}{sin[ \pi(x + 1)]}\)

or

\(\displaystyle \lim_{x \to -1} \frac{ \sqrt[3]{x} + 1}{sin( \pi(x + 1))}\)

or

\(\displaystyle \lim_{x \to -1} \frac{ \sqrt[3]{x} + 1}{sin( \pi[x + 1])}\)

or etc.

 

[wolly]

\(\displaystyle \lim_{x \to -1} \frac{ \sqrt[3]{x} + 1}{sin[ \pi(x + 1)]}\)

or

\(\displaystyle \lim_{x \to -1} \frac{ \sqrt[3]{x} + 1}{sin( \pi(x + 1))}\)

or

\(\displaystyle \lim_{x \to -1} \frac{ \sqrt[3]{x} + 1}{sin( \pi[x + 1])}\)

or etc.

The first one

 

[Steven G]

Wolly, lookagain is saying that all three that he listed are ALL the same.
The say that you wrote it, you can simply bring \(\displaystyle \sin\pi\) in front of the limit. Do you see that?

 

[wolly]

I solved IT using subtitution and I got 1/3π using both methods!

 

[mario99]

I solved IT using subtitution and I got 1/3π using both methods!

Correct answer.

[math]x=t-1[/math]How did this answer appeared?

Did you know why they used [imath]x = t - 1[/imath]?

 

[lookagain]

I solved IT using subtitution and I got 1/3π ...

In horizontal form, you need grouping symbols for that denominator: \(\displaystyle \ 1/(3 \pi ). \)