Substitute digits in place of letters

So obviously this can be solved by brute force–trying different values for each letter.  The challenge is how to deduce the answer with as little trial and error as possible.  We can narrow it down with some observations.

Both I’s in LIMITS line up under an I, so let’s start there.
Because S can’t be zero and S + I results in I as the last digit, we know T + T >= 9 (to generate carry) and that S = 9.
Since S = 9, that must mean that H = 0 so that H + I = I with no carry.
Since S = 9, L + N = 9.
There is no carry from L + N, so L + I = T or 10 + T, but we know there is also no carry from L + I and that I is not zero.
Because T + T = 10 + M and M can’t be zero, then T isn’t 5.
Because WITHIN and LIMITS are both six digits, we know that L = W + 1.

So before any trial and error we have:

S = 9
H = 0
T = {6,7,8} = L + I
M = {2,4,6} = 2*T – 10
W = {1,2,3,4,5,6} = L – 1
L = {2,3,4,5,6,7} = W + 1 = 9 – N
N = {2,3,4,5,6,7} = 9 – L
I = {1,2,3,4,5,6} = T – L

Looking at this, we see that L has a relationship to every other letter, so let’s look at that.  Since W = L – 1 and N = 9 – L, L can’t be 5 since that would require both W and N to be 4.  So now the possibilities are:

S = 9
H = 0
T = {6,7,8} = L + I
M = {2,4,6} = 2*T – 10
W = {1,2,3,5,6} = L – 1
L = {2,3,4,6,7} = W + 1 = 9 – N
N = {2,3,5,6,7} = 9 – L
I = {1,2,3,4,5,6} = T – L

Now we’re ready for trial and error.  Since there is only one way to assign 8 and since eight out of ten digits are used (only two unused), try assigning T = 8, which gives M = 6.  Since 6 is taken, neither L nor I can be 2 (since 8 – 6 = 2) nor 4 (since they need to be different and sum to 8).  Also, since L = 9 – N and N can’t be 6, then L can’t be 3 and L = W + 1 and W can’t be 6, so L can’t be 7, leaving no possible value for L.  So we know T can’t be 8, leaving:

S = 9
H = 0
T = {6,7} = L + I
M = {2,4} = 2*T – 10
W = {1,2,3,5} = L – 1
L = {2,3,4,6} = W + 1 = 9 – N
N = {3,5,6,7} = 9 – L
I = {1,2,3,4,5} = T – L

There are only two ways to assign 7, so try assigning T = 7, which gives M = 4.  Since 4 is taken, neither L nor I can be 3 or 4.  Since N can’t be 7 and N = 9 – L, L also can’t be 2, leaving only L = 6:

S = 9
H = 0
T = 7
M = 4
W = 5
L = 6
N =3
I = 1

This give a unique assignment for each letter, so see if it works:  97166 + 517013 = 614179
It does work, on only the second (of three possible) guesses.