L'Hospital problem?

Question

I am stuck on these 2 problems. How would you use L'hospitals rule to solve for:

1) lim as x->0+ (((3x +1)/x) - (1/sinx))
and
2)lim as x-> 0 (e^x + x)^(1/x)

Any help would be much appreciated. Who ever knew calculus could be so tough. Lol.

Answer 1

l'Hôpital's rule: if lim (x->c) f(c)/g(c) becomes 0 or infinity, then:
lim (x->c) f(c)/g(c) = lim (x->c) f'(c)/g'(c).

So:
1.) lim (x->0) (((3x +1)/x) - (1/sinx))
f(x) = 3x + 1, g(x) = x
h(x) = 1, i(x) = sin x
so: lim (x->0) f(x)/g(x) - h(x)/i(x) = lim (x->0) f'(x)/g'(x) - h'(x)/i'(x)

Now, find the derivatives for all 4 functions listed above:
f'(x) = 3, g'(x) = 1
h'(x) = 0, i'(x) = cos x

Giving you:
lim (x->0) f'(x)/g'(x) - h'(x)/i'(x) = 3/1 - 0 / cos x = 3 - 0/1 = 3 (solution!)

For #2, you need to differentiate so that the exponent doesn't have just x in the denominator.