2
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Recently I got in a problem and I will try to explain with a simple example. I have an expression which reads :

y = (10 + 2*x )/x;

Now I want to substitute 1/x -> 0 . Now if I substitute directly in this expression the result is zero i.e.

y = (10 + 2*x )/x;
y /. {1/x -> 0};
Print[y];

Answer = 0 which should not be.

So I have to expand before substituting i.e.

y = (10 + 2*x )/x;
y1=Expand[y,x];
y1 /. {1/x -> 0};
Print[y1];

which gives me the correct answer = 2.

My question is, is there any efficient way (or another way) to do this work. This is because in case of 'y' has a very large expression or coefficient of 1/x is very complicated, Expand command is taking very very long time.

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6
  • 1
    $\begingroup$ What about taking the limit x->Infinity ? $\endgroup$
    – Mher
    Jun 1, 2016 at 12:37
  • $\begingroup$ Hi @Mher Safaryan , if you are suggesting not to use Expand and just to use y /. {x ->Infinity} , it won't work. The problem will be the same no? Previously it gave 0, now it will give Infinity. $\endgroup$
    – BabaYaga
    Jun 1, 2016 at 12:43
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    $\begingroup$ Why it is infinity ? it's just 2 for your example. Just try this (* Limit[y, {x -> Infinity}] *) $\endgroup$
    – Mher
    Jun 1, 2016 at 12:50
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    $\begingroup$ I believe that @Mher was suggesting that you use Limit[y, x -> Infinity], which does yield 2. So does First@Collect[y, x] or Coefficient[y, x, 0]. The last may be best, depending on your goal. $\endgroup$
    – bbgodfrey
    Jun 1, 2016 at 13:04
  • $\begingroup$ okay both are working .. thank you both. $\endgroup$
    – BabaYaga
    Jun 1, 2016 at 13:27

1 Answer 1

1
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If you want to substitute 1/x->0 in in expressions then take the limit x->Infinity:

y = (10+2*x)/x;
Limit[y, x->Infinity]

Other ways of doing that is the followings (see bbgodfrey's comment):

First@Collect[y, x]
Coefficient[y, x, 0]
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