Finding limits using L'Hopital's Rule?

Question

a) lim cotx/lnx
x->0+

b) lim (1-3/x)^x
x-> +oo

Please help me solve these and also show the steps on how to do these problems. Thanks

Answer 1

#a:
[x→0+]lim cot x/ln x gives ∞/-∞, so take the derivative of both the top and the bottom:

[x→0+]lim -csc² x/(1/x) = [x→0+]lim -x csc² x, which yields 0*∞, so rewrite csc² x as 1/sin² x to get the form x/sin² x, which reduces to 0/0, and apply L'hopital's rule a second time:

[x→0+]lim -1/(2 sin x cos x), which is -1/0. At this point, the limit is determinate, and since we are approaching from the positive side, the denominator is always positive, so this limit is -∞

#b: This limit is substantially easier. First, instead of trying to find [x→∞]lim (1-3/x)^x directly, we shall instead try to find ln [x→∞]lim (1-3/x)^x. Since the logarithm is continuous, we can transpose it with the limit to obtain:

[x→∞]lim ln (1-3/x)^x

Using the laws of logarithms:

[x→∞]lim x ln (1-3/x)

This gives the indeterminate form ∞*0. So We rewrite x as 1/(1/x) to obtain:

[x→∞]lim ln (1-3/x)/(1/x)

Applying L'hopital:

[x→∞]lim 1/(1-3/x) * (3/x²)/(-1/x²)

Simplifying:

[x→∞]lim -3/(1-3/x)

Which is clearly -3. But this establishes that ln [x→∞]lim (1-3/x)^x = -3. To find the limit itself, exponentiate both sides to obtain [x→∞]lim (1-3/x)^x = e^(-3).

Answer 2

No. Figure it out yourself.