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### Limit Proofs with L'Hopital's Rule

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Date: 05/27/98 at 10:28:31
From: Kumarpal Sheth
Subject: Proofs for limits

Hello Dr. Math,

I have been just introduced to calculus. In limits, we have the
following identities as results without proof:

1)  lim   (x + 1/x)^x = e
x-->0

2)  lim   (a^x - 1)/x = ln(a)
x-->0

3)  lim   (ln(1+x))/x = 1
x-->0
```

```
Date: 05/27/98 at 13:26:12
From: Doctor Rob
Subject: Re: Proofs for limits

Later on in your study of calculus you will learn something called
L'Ho^pital's Rule. This will allow you to compute and prove these
limits. It says:

Theorem:  If as x -> a (where a is any real number or infinity),
lim f(x) = 0 and lim g(x) = 0, then:

lim [f(x)/g(x)] = lim [f'(x)/g'(x)],
x->a              x->a

provided either limit exists (in which case both do).

Corollary: If lim f(x) = infinity and lim g(x) = infinity, then:

lim [f(x)/g(x)] = lim [f'(x)/g'(x)],
x->a              x->a

provided either limit exists (in which case both do).

Proof of Corollary:

lim f(x)/g(x) = lim [1/g(x)]/[1/f(x)]
x->a              x->a

and we can apply the Theorem to this quotient:

lim f(x)/g(x) = lim [-g'(x)/g(x)^2]/(-f'(x)/f(x)^2]
x->a            x->a

= {lim [g'(x)]/f'(x)]}*{lim [f(x)/g(x)]}^2
x->a                 x->a

1/lim [g'(x)/f'(x)] = lim [f(x)/g(x)]
x->a                x->a

lim [f(x)/g(x)] = lim [f'(x)/g'(x)]               Q.E.D.
x->a              x->a

If you accept this theorem without proof, the above three limits can
be computed using it:

1) lim (x + 1/x)^x = lim e^(x * ln[x + 1/x])
x->0              x->0

= e^lim ln[x + 1/x]/(1/x)
x->0

Now the limit has the form lim f(x)/g(x), where f(x) = ln(x + 1/x)
and g(x) = 1/x, and lim f(x) = infinity, and lim g(x) = infinity as
x -> 0. Apply the corollary to L'Ho^pital's Rule:

f'(x) = (x^2 - 1)/[x * (x^2 + 1)]
g'(x) = -1/x^2
f'(x)/g'(x) = -x * (x^2 - 1)/(x^2 + 1)

The limit of this as x -> 0 is 0. Thus:

lim (x + 1/x)^x = e^lim -x*(x^2-1)/(x^2+1)
x->0                x->0

= e^0
= 1

2) lim (a^x - 1)/x
x->0

This is already in the correct form for L'Ho^pital's Rule, with
f(x) = a^x - 1 and g(x) = x. The hard part here is computing f'(x),
but a = e^ln(a), so a^x = e^(x*ln(a)), and:

f'(x) = ln(a)*e^(x*ln(a)) = a^x*ln(a)

g'(x) = 1

Putting this together:

lim (a^x - 1)/x = lim a^x*ln(a)/1 = ln(a)
x->0              x->0

3) lim [ln(1 + x)]/x = lim [1/(x + 1)]/1 = 1
x->0                x->0

If you want a proof of L'Ho^pital's Rule, either wait until you get to
that point in calculus, or write again.

-Doctor Rob,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```

```
Date: 05/27/98 at 15:18:34
From: Kumarpal Sheth
Subject: Re: Proofs for limits

Hello Dr. Math,

Thanks a lot for the proof. I didn't understand everything but I will
try and figure it out. I'll let you know if I have further questions.

Thanks a lot once again.

Kumarpal Sheth
```
Associated Topics:
High School Calculus

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