Someone help me with limits?

Question

lim [log(1-x) + x*e^(x/2)] / x^3

x->0

and

lim log(1+x^2) / 2x

x-->0

Answer 1

Both of your problems are in the indeterminate form 0/0. So you must use L'Hospital's Rule

Take derivatives of numerator and denominator separately. Then see if a limit results. If it is still 0/0 repeat the process. At some point you will arrive at a denominator of 6 and at this time you can determine the limit.

Answer 2

Just use L'Hopital to compute these limits:

1)

lim [log(1-x) + x*e^(x/2)] / x^3 =

x->0

lim [1/(x-1) + e^(x/2) + (x/2)*e^(x/2)] / 3x^2 =

x->0

lim [-1/(x-1)^2 + e^(x/2) + (x/4)*e^(x/2)] / 6x =

x->0

lim [2/(x-1)^3 + (3/4)e^(x/2) + (x/8)*e^(x/2)] / 6

x->0

But this limit isn't indeterminate anymore, and since the function inside the limit is continuous at x=0, we can plug in 0 to find that the initial limit equals (-2 + 3/4)/6 = -5/24

(There might be a faster way to do number 1 with Taylor series, but I didn't see it immediately.)

2)

lim log(1+x^2) / 2x =

x->0

lim x/(1+x^2) =

x->0

0

Answer 3

because the numerator and denominator become zero when x->, u can use the L'Hopital rule by taking the derivative of each separately. Do this until u dont get a "0/0, or infinity/infinitiy"