How Do You Resolve a Division by Zero in Limit Calculations?

[QuarkCharmer]

Homework Statement

\lim_{t\to0}\frac{1}{t}-\frac{1}{t^2+t}

Homework Equations

Limit Laws, et al.

The Attempt at a Solution

\lim_{t\to0}\frac{1}{t}-\frac{1}{t^2+t}

I didn't feel this would work, too much division by zero, so I found a common den.

\lim_{t\to0}\frac{t^2+t}{t(t^2+t)}-\frac{t}{t(t^2+t)}

\lim_{t\to0}\frac{t^2+t-t}{t(t^2+t)}

\lim_{t\to0}\frac{t^2}{t^3+t^2)}

\lim_{t\to0}\frac{t^2}{t^2(t+1)}

\lim_{t\to0}\frac{1}{t+1} = \frac{1}{1}

Edit:
I figured it out now, does it appear correct?

 

[JNBirDy]

You *can* use http://mathworld.wolfram.com/LHospitalsRule.html"

=lim(t->0) t²/t³+t²

=lim(t->0) 2t/3t²+2t

=lim(t->0) 2/3t+2

=2/2

=1 (since 3t -> 0)

 

[QuarkCharmer]

Well, I arrived at 1 without the rule? We have not covered L'Hospital so I know I should be able to complete it without. Have I made a mistake?

 

[Dick]

You need to use http://mathworld.wolfram.com/LHospitalsRule.html"

=lim(t->0) t²/t³+t²

=lim(t->0) 2t/3t²+2t

=lim(t->0) 2/3t+2

=2/2

=1 (since 3t -> 0)

You don't NEED to use l'Hopital. Reducing the problem using algebra to the determinate form 1/(t+1) does the job as well.

 

[JNBirDy]

Well, I arrived at 1 without the rule? We have not covered L'Hospital so I know I should be able to complete it without. Have I made a mistake?

The way you did it seems to work as well. L'Hospital's Rule is still nifty to know since it will let you check your work.

You don't NEED to use l'Hopital. Reducing the problem using algebra to the determinate form 1/(t+1) does the job as well.

Yeah I was thinking of editing out the word 'need' after I replied but was too lazy. :)

 

[QuarkCharmer]

Thanks guys.

 

[TROGLIO]

1/t - 1/t^2+t
add and substract t in d numerator of -1/t^2+t

 

[HallsofIvy]

Homework Statement

\lim_{t\to0}\frac{1}{t}-\frac{1}{t^2+t}

Homework Equations

Limit Laws, et al.

The Attempt at a Solution

\lim_{t\to0}\frac{1}{t}-\frac{1}{t^2+t}

I didn't feel this would work, too much division by zero, so I found a common den.

\lim_{t\to0}\frac{t^2+t}{t(t^2+t)}-\frac{t}{t(t^2+t)}

Better is the least common denominator- t^2+ 1= t(t+ 1) so t(t+ 1) is itself a common denominator.

(t+1)/(t(t+1))- \frac{1}{t(t+1)}= t/(t(t+1))= 1/(t+ 1).

Now take the limit.

\lim_{t\to0}\frac{t^2+t-t}{t(t^2+t)}

\lim_{t\to0}\frac{t^2}{t^3+t^2)}

\lim_{t\to0}\frac{t^2}{t^2(t+1)}

\lim_{t\to0}\frac{1}{t+1} = \frac{1}{1}

Edit:
I figured it out now, does it appear correct?

 

[Mark44]

The way you did it seems to work as well. L'Hospital's Rule is still nifty to know since it will let you check your work.

But since QuarkCharmer hasn't learned this rule yet, it's not an option in this problem. Also, there are limits for which L'Hopital's Rule can be used, but it doesn't lead to finding the limit. For problems such as these, being able to simplify using algebra is the way to go.

Yeah I was thinking of editing out the word 'need' after I replied but was too lazy. :)

 

[hunt_mat]

Why not just write:
<br /> \frac{1}{t}-\frac{1}{t^{2}+t}=\frac{1}{t}\left( 1-\frac{1}{t+1}\right) =\frac{1}{t}\cdot\frac{t}{t+1}=\frac{1}{t+1}<br />

 

[HallsofIvy]

By the way

1) You don't divide by 0. The fact that attempting to set t= 0 in this fraction gives a 0 in the denominator means you can't find the limit that way.

2) While it isn't necessary to say it every time, you should keep in mind that
\frac{t}{t(t+1)}= \frac{1}{t+1}
is only true for t\ne 1. Fortunately, if f(x)= g(x) for all x except a, lim_{x->a} f(x)= lim_{x\to a} g(x) so we don't need to worry about that.