Tackling a Tricky Limit: 0/0 & ∞-∞

[mtayab1994]

Homework Statement

Calculate the following limit:

\lim_{x\rightarrow0}\frac{1}{x^{2}}-\frac{1}{xsinx}

The Attempt at a Solution

I've tried everything possible and i keep getting undefined forms like 0/0 and ∞-∞. Any ideas anyone??

 

[DryRun]

It is indeed an indeterminate form and you should use L'Hopital's Rule:
\lim_{x\rightarrow0}\frac{1}{x^{2}}-\frac{1}{xsinx}=\lim_{x\rightarrow0}\frac{\sin x-x}{x^2 \sin x}=\frac{0}{0}

 

[SammyS]

Homework Statement

Calculate the following limit:

\lim_{x\rightarrow0}\frac{1}{x^{2}}-\frac{1}{xsinx}

The Attempt at a Solution

I've tried everything possible and i keep getting undefined forms like 0/0 and ∞-∞. Any ideas anyone??

Can you list what you've tried in more detail so we don't go around in circles?

 

[mtayab1994]

Can you list what you've tried in more detail so we don't go around in circles?

I've tried getting a common denominator like sharks stated, and then i tried using trig rules which i couldn't and then now I'm stuck.

 

[mtayab1994]

I was able to solve it easily using l'hopital's rule 3 times repeatedly and i got an answer of -1/6, but doing it without l'hopital's rule is really difficult.

 

[Steely Dan]

I was able to solve it easily using l'hopital's rule 3 times repeatedly and i got an answer of -1/6, but doing it without l'hopital's rule is really difficult.

Using the Taylor series expansion for sin(x) isn't difficult...if you have an indeterminate form limit, trying to algebraically shuffle your way out of it won't help you very much.

 

[DryRun]

I was able to solve it easily using l'hopital's rule 3 times repeatedly and i got an answer of -1/6, but doing it without l'hopital's rule is really difficult.

3 times? It should be only once.

 

[mtayab1994]

3 times? It should be only once.

No, It took me 3 times, because I didn't use taylor expansion.

 

[DryRun]

\lim_{x\rightarrow0}\frac{\sin x-x}{x^2 \sin x}=\lim_{x\rightarrow0}\frac{\cos x-1}{2x \sin x+ x^2\cos x}=\frac{0}{0}

 

[gopher_p]

\lim_{x\rightarrow0}\frac{\sin x-x}{x^2 \sin x}=\lim_{x\rightarrow0}\frac{\cos x-1}{2x \sin x+ x^2\cos x}=\frac{0}{0}

... which is indeterminate.

 

[DryRun]

The answer is: -\frac{1}{6}. You are correct. But i don't see any other method.

 

[Karamata]

Well, you know that \displaystyle\lim_{x\to 0}\frac{1}{x^2}-\frac{1}{x\sin x} = -\frac{1}{6} (using l'hopital's rule), but, you want to prove it without using l'hopital's rule, so you prove it by definition ((ε,δ)-definition of limit)

so, you use l'hopital's rule to see what is the limit, and then, you prove it by definition (without l'hopital's rule)

sorry for bad English

 

[scurty]

Does the problem ask you to solve the limit without taylor series or l'Hospital's rule, or are you just wondering if it can be solved without those methods?

 

[mtayab1994]

Does the problem ask you to solve the limit without taylor series or l'Hospital's rule, or are you just wondering if it can be solved without those methods?

No I'm just wondering if it could be solved without them.

 

[mtayab1994]

Well, you know that \displaystyle\lim_{x\to 0}\frac{1}{x^2}-\frac{1}{x\sin x} = -\frac{1}{6} (using l'hopital's rule), but, you want to prove it without using l'hopital's rule, so you prove it by definition ((ε,δ)-definition of limit)

so, you use l'hopital's rule to see what is the limit, and then, you prove it by definition (without l'hopital's rule)

sorry for bad English

I don't want to prove the limit i just want to solve it.

 

[mtayab1994]

\lim_{x\rightarrow0}\frac{\sin x-x}{x^2 \sin x}=\lim_{x\rightarrow0}\frac{\cos x-1}{2x \sin x+ x^2\cos x}=\frac{0}{0}

And then derive what you have 2 more times and you'll get -cos(x)/something really long.

Then when you compute it you will get -1/6, but I don't get how you did it using l'hospital's rule once only??

 

[Curious3141]

The easiest way is to first bring everything to a common denominator, then use the Taylor series for sin(x) to the x^3 term on both numerator and denominator. Simplify, neglect higher order terms and it gives you a quick answer.