Expert: Scotto - 7/2/2009

Question
How do you use L'Hospital's Rule to compute
1.  Lim as x approaches pos 0 of (cos x - e^x)/(ln(1+x^2)
2.  lim as x approaches 0 of (x/sin x)^(1/x^2)

Answer
1.  Lim as x->+0 of (cos x - e^x)/(ln(1+x^2)

L'Hospital's Rule says to take the derivative of the numerator and the denominator, and then see what the quotient comes out to be.

The derivative of the numerator is -sinx - e^x.
The lim(x->0) (-sinx - e^x) is -sin(0) - e^0 = -0 - 1 = -1.

The derivative of the denominator is 2x/(1+x^2).
The lim(x->0)(2x/(1+x^2)) is 2*0/(1+0^2) = 0/1 = 0.

Using this, the quotient is -1/0 = -∞, so the
lim as x->+0 of (cos x - e^x)/(ln(1+x^2) = -1/0 = -∞.


2.  lim as x approaches 0 of (x/sin x)^(1/x^2)

lim(x->0)((x/sinx)^(1/x²) = lim(x->0)(e^ln((x/sinx)^(1/x²)))
= e^lim(x->0)(ln((x/sinx)^(1/x²)))
= e^lim(x->0)((1/x²)ln(x/sinx))
= e^lim(x->0)(ln(x/sinx)/x²).

For the numerator, lim(x->0)(ln(x/sinx)) = ln(lim(x->0)(x/sinx)).
It is known that the lim(x->0)(sinx/x) is 1, and we have the
lim(x->0)(x/sinx).  This is the same as 1/lim(x->0)(sinx/x), so the answer is 1/1 = 1.

In case this needs to be shown, apply L'Hospital's Rule to the top.
The numerator is x and the denominator is sinx.  The derivative of the numerator is 1 and the derivative of the denominator is cosx.  As x -> 0, the numerator goes to 1 and the denominator goes to 1, so the limit is 1/1, which is 1.

For the denominator, the lim(x->0)(x²) = 0.

Now putting the answers to these limits back into the equation gives
e^(1/0) = ∞.





Humor: So if the first problem is -∞ and the second problem is ∞,
then the sum of the two might be 0.  Does this mean this question is asking for nothing at all...