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Consider

$\qquad \underset{h\to 0}{\text{lim}}\dfrac{\sqrt[4]{h+x}-\sqrt[4]{x}}{\sqrt{h+x}-\sqrt{x}}$

Here is what I get when I evaluate it Mathematica:

MMA output

When I calculate it by hand and in other applications, it results in

$\qquad \dfrac{1}{ 2 \sqrt[4]{x}}$

Syntax error in Mathematica?

Update

It is not that I do not know how to calculate the limit, I have easily done it by hand without derivatives and with the rule of L'hopital. I find it strange, that when writing it using one of the assistant palettes, it evaluated to Indeterminate. I imagined that I had to put an extra condition.

That's why I thought maybe I needed a condition for x , let's say x > 0, but I didn't know how to write it.

This result threw me. Now, I did it again and it I got the correct result (maybe some variable in memory was bothering me).

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    $\begingroup$ This limit is immediatelly calculated by hand. You should not use Mathematica to figure it out. Nethertheless take a look at How can I calculate the limit without using L'Hôpital's rule $\endgroup$
    – Artes
    Nov 4, 2020 at 17:54
  • $\begingroup$ @Artes_see update above $\endgroup$
    – BeTDa
    Nov 4, 2020 at 23:35
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    $\begingroup$ Possibly you had set x to 0? $\endgroup$
    – Carl Woll
    Nov 4, 2020 at 23:50
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    $\begingroup$ I’m voting to close this question because no one can reproduce the problem the OP is experiencing; I suspect it is being caused by corrupted lexical elements in the OP's Mathematica notebook. $\endgroup$
    – m_goldberg
    Nov 5, 2020 at 2:21
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    $\begingroup$ This Question has no question. $\endgroup$ Nov 5, 2020 at 5:59

2 Answers 2

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Responding to the only question-like feature of the edited Question,

Assuming[{x > 0},
    Limit[ /* more stuff here */ ]
]
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  • $\begingroup$ ,That was missing, I thought that the limit was not working for me because the "x" was not defined.Thanks $\endgroup$
    – BeTDa
    Nov 5, 2020 at 16:41
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Mathematica

Limit[((h + x)^(1/4) - ( x)^(1/4))/((h + x)^(1/2) - ( x)^(1/2)),h -> 0]
(*1/(2 x^(1/4))*)

evaluates the limit without problems

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  • $\begingroup$ see update above $\endgroup$
    – BeTDa
    Nov 4, 2020 at 23:35
  • $\begingroup$ @BeTDa It's still unclear what you tried. Please provide Mathematica code of your input "In[27]" .Thanks! $\endgroup$ Nov 5, 2020 at 7:16
  • $\begingroup$ Ulrich Neumann That's all the code, I wrote it using the MMA's teaching assistant, surely it was set in memory at some point x=0, as I was told. At that time and several times later I always got a long time, I thought about putting a restriction, but I did not know how to do it $\endgroup$
    – BeTDa
    Nov 5, 2020 at 16:39
  • $\begingroup$ @BeTDa A pixel image can't be "all the code" , that's my question $\endgroup$ Nov 6, 2020 at 7:56

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