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### L'Hopital's Rule

```
Date: 23 Apr 1995 01:24:03 -0400
From: Naseema Nabi
Subject: HELP!

Dear Dr. Math,

Help me, I can't figure out how to take the limit using L'Hopital's rule
on this one problem.

Lim [(arctan X - (3.14(Pie)/4)/(X-1)]
x=1

```

```
Date: 23 Apr 1995 12:49:56 -0400
From: Dr. Sydney
Subject: Re: HELP!

Dear Naseema,

Hello!  I think on this problem you actually don't need to use L'Hopital's
Rule.  However, I am a bit unclear on what the problem is.  I think the way
you have written your parentheses the problem is this:

lim  arctan x - pi/4(x-1)
x->1

(So, the x-1 is in the denominator)

In this case, think about what happens to the last term (the pi/4(x-1) term)
as x approaches one.  The closer you get to one, the smaller (x-1) becomes,
and the bigger pi/4(x-1) becomes, right? (Note:  this is in the case that
you are approaching one from numbers bigger than one.  Can you see what
happens if you approach one by numbers smaller than one?).  So, there is no
limit as x approaches one.

If, however, the problem is

lim arctan x - pi(x-1)/4
x->1

Then, split the limit up into the sum of two limits, and so you have

lim arctan x - lim pi(x-1)/4
x->1           x->1

Evaluate these two limits (you still won't need to L'hop...) and you'll have

If any of this is unclear, or if you have any other questions feel free to
write back!

--Sydney, "Dr. Math!"
```

```
Date: 23 Apr 1995 15:04:43 -0400
From: Dr. Ken
Subject: Re: HELP!

Hello there!

I'm another math doctor here, and I interpreted your message in yet another
way.  It looks to me like you have

Arctan(x) - Pi/4
Lim x->1  ------------------
x-1

Is this right?  Then you _would_ use L'Hopital's rule; let's review the
criteria for when you can use L'Hopital.  Use it when the numerator and the
denominator are both going to positive infinity, or when they're both going
to zero, or when they're both going to negative infinity. In this case, if
you plug in 1 for x in the to and the bottom, you'll get zero.  So Yippee,
we can use L'Hopital!  That means we can take the derivative of the top and
bottom.

The derivative of the top is 1/(1 + x^2), and the derivative of the bottom
is 1.  So we have

1          1      1
Lim x->1  --------- = ------- = - .
1 + x^2     1 + 1    2

I hope this helps!

-Ken Dr. Math
```
Associated Topics:
High School Calculus

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