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I need to prove this limit without using the L'Hopital's rule:

$$\lim_{x\to 0} \frac{(1+a\,x)^{1/4} - (1+b\,x)^{1/4}}{x} = \frac{a-b}{4}$$

How can I do it in Mathematica?

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I have decided not to close this question or migrate, because it has a sensible Mathematica-based answer. As such it will still be useful to the community, even if the OP wasn't actually wanting a Mathematica solution. There is only one more vote needed to close, and I'm happy to be overruled. –  Verbeia Apr 14 at 23:24
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2 Answers 2

One can simply use:

Limit[((1 + a x)^(1/4) - (1 + b x)^(1/4))/x, x -> 0]

to get the result. However Limit does know about the l'Hopital rule.
Nonetheless there are different ways to go, e.g. let's use Series to write down the first few terms of the Taylor series of ((1 + a x)^(1/4) - (1 + b x)^(1/4))/x:

Series[((1 + a x)^(1/4) - (1 + b x)^(1/4))/x, {x, 0, 3}]
 (a - b)/4 - 3/32 (a^2 - b^2) x + 7/128 (a^3 - b^3) x^2 
  - (77 (a^4 - b^4) x^3)/2048 + O[x]^4

This demonstrates clearly what the limit of the expression is when x -> 0. Unfortunately this still uses some analytical methods and it might be especially useful if we can get rid of them.

Elementary Proof

Therefore we can provide quite an elementary proof (only ancient Greeks' methods). We take the expression ((1 + a x)^(1/4) - (1 + b x)^(1/4)) and multiply it by

expr1 = ((1 + a x)^(1/4) + (1 + b x)^(1/4));

The latter term is equal to 2 at x == 0 (expr1 /. x -> 0 yields 2).

((1 + a x)^(1/4) - (1 + b x)^(1/4)) ((1 + a x)^(1/4) + (1 + b x)^(1/4)) // Expand
Sqrt[1 + a x] - Sqrt[1 + b x]

Now we take Sqrt[1 + a x] - Sqrt[1 + b x] and we multiply it by

expr2 = Sqrt[1 + a x] + Sqrt[1 + b x];

similarily the latter term is equal to 2 at x == 0. Now we have:

(Sqrt[1 + a x] - Sqrt[1 + b x]) (Sqrt[1 + a x] + Sqrt[1 + b x]) // Expand
a x - b x

We have multiplied the interesting term twice by 2 and now we take the denominator (i.e. x) of the initial exppession. Thus we can see from the last output that $$\lim_{x\rightarrow 0 }\frac{(1 + a x)^{\frac{1}{4}} - (1 + b x)^{\frac{1}{4}}}{x}=\frac{a-b}{4}$$ Q.E.D.

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I am guessing this is math homework and should be migrated, don't you think so? –  sebhofer Apr 14 at 23:04
    
Not necessarily, I demonstrated what the OP wanted without the l'Hopital rule. I could also provide another ways. Just showed something for those about to study calculus, with the help of Mathematica. Right? –  Artes Apr 14 at 23:08
    
Ok, we will see what the OP says anyways. –  sebhofer Apr 14 at 23:17
    
i'm sure its trivial to show that the series converges, but i'd think you need to show that step to offer this up as proof. –  george2079 Apr 14 at 23:30
    
@george2079 Now you have got what you wanted. –  Artes Apr 15 at 0:01
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First, let us notice that the limit follows from the following:

Lemma. $\displaystyle \lim_{u \rightarrow 0} {(1+u)^{1/4}-1 \over u} = {1 \over 4}$.

For

$${{(1+a\,x)^{1/4} - (1+b\,x)^{1/4}} \over {x}} = {{(1+a\,x)^{1/4} - 1} \over {x}} - {{(1+b\,x)^{1/4} - 1} \over {x}}$$ $$ = a\,{{(1+u)^{1/4} - 1} \over {u}} - b\,{{(1+v)^{1/4} - 1} \over {v}}\,,$$

where $u = ax$ and $v = bx$.

Mathematica proofs

Sans L'Hôpital:

The nub of a proof is an equality like this:

Simplify[Abs[((1 + u)^(1/4) - 1)/u - 1/4] < Abs[u], -1 < u < 1 && u != 0]
(* True *)

More directly from the definition:

Resolve[ForAll[epsilon, 
  Implies[epsilon > 0, 
   Exists[delta, 
    ForAll[u, 
     Implies[Abs[u] < delta && u != 0, 
      Abs[((1 + u)^(1/4) - 1)/u - 1/4] < epsilon]]]]],
 Reals
 ]
(* True *)

Another elementary proof

Proof á la Euclid. [Prompted by @Artes's remark about ancient Greeks.]

Mathematica graphics

Let $BX$ be given with $OA=OB=1$ and $AX = u>0$. Let $OP$ have been drawn perpendicular to $BX$ with $OC=1$. Let $BX$ have been bisected at $Q$. With center $Q$ and distance $QX$, let circle $XPB$ have been described. Let $PQ$ be joined. Then $PQ = BQ = 1+u/2$ and $OP = \sqrt{1+u}$; further $OP < PQ = BQ$. Subtracting $BO = OC = 1$ yields $ CP = \sqrt{1+u}-1 < OQ = u/2$.

On the other hand, let $QM$, $XH$ have been drawn perpendicular to $OX$, and let $CG$ and $PH$ have been drawn perpendicular to $OP$. Let $PX$ have been joined and intersect $QM$ at $K$. Finally let a line have been drawn perpendicular to $QM$ at $K$. The complements $OK$, $KH$ of the diagonal $PX$ are equal [Eucl. I.43]. Therefore the rectangle $CH$ is greater than the rectangle $OK$ which is greater than $OG$. Thus $CH = (1+u)(\sqrt{1+u}-1) > u/2$.

Therefore $${u \over 2(1+u)} < \sqrt{1+u}-1 < {u \over 2}$$ or $$1 + {u \over 2(1+u)} < \sqrt{1+u} < 1 + {u \over 2}$$

Similarly, letting $OX = \sqrt{1+u}$, we obtain $$1 + { \sqrt{1+u}-1 \over 2( \sqrt{1+u})} < (1+u)^{1/4} < 1 + {\sqrt{1+u}-1 \over 2}$$ Applying the previous inequalities, we have $$1 + {u \over 4(1+u/2)} < 1 + { \sqrt{1+u}-1 \over 2( \sqrt{1+u})} \quad\hbox{and}\quad 1 + {\sqrt{1+u}-1 \over 2} < 1 + {u \over 4}\,.$$ From this we get $$1 + {u \over 4(1+u/2)} < (1+u)^{1/4} < 1 + {u \over 4}\,.$$ It follows that $$\left| {(1+u)^{1/4} -1 \over u} - {1 \over 4} \right| < {u \over 8 + 4u} < u\,.$$

Similarly, taking $X$ between $O$ and $A$ so that $AX = -u$, one can show that $$\left| {(1+u)^{1/4} -1 \over u} - {1 \over 4} \right| < {-u}\,,$$ provided $-1/2 < u < 0$. Thus the lemma is established.

Calculus proof

The limit in the lemma is the derivative of $x^{1/4}$ at $x = 1$.


Code dump for figure

labels[u_] := MapThread[Text, Transpose[{
     {"O", {0, 0}, {0, 1.5}},
     {"A", {1, 0}, {0, 1.5}},
     {"B", {-1, 0}, {-1.8, 1.5}},
     {"C", {0, 1}, {1.5, 0}},
     {"X", {1 + u, 0}, {-1.8, 1.5}},
     {"P", {0, Sqrt[1 + u]}, {1.5, -1}},
     {"Q", {u/2, 0}, {0, 1.5}},
     {"G", {u/2, 1}, {1.5, 1.5}},
     {"H", {1 + u, Sqrt[1 + u]}, {-1.8, -1}},
     {"K", {u/2, (2 + u)/2/Sqrt[1 + u]}, {-2, -1.5}},
     {"L", {u/2, 1}, {-2.7, 1.5}},
     {"M", {u/2, Sqrt[1 + u]}, {0., -2.5}}
     }]];
Manipulate[
 With[{u = Exp[logu]},
  Graphics[
   {Point[{{0, 0}, {1, 0}, {0, 1}, {-1, 0},
      {1 + u, 0}, {u/2, 0}, {0, Sqrt[1 + u]}}],
    Circle[{u/2, 0}, 1 + u/2],
    Line[{{{-1, 0}, {1 + u, 0}, {1 + u, Sqrt[1 + u]}, {0, Sqrt[
        1 + u]}, {0, 0}}, {{0, 1}, {1 + u, 1}}, {{u/2, 0}, {u/2, Sqrt[
        1 + u]}}, {{u/2, 0}, {0, Sqrt[1 + u]}, {1 + u, 
        0}}, {{0, (2 + u)/2/Sqrt[1 + u]}, {1 + u, (2 + u)/2/Sqrt[
         1 + u]}}}],
    labels[u]
    },
   PlotRange -> {{-1.1, 2.1}, {-0.5, 1.5}},
   ImageSize -> 400
   ]
  ],
 {{logu, -0.3}, -3., 0.}
 ]
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To suggest correctness of the stated Lemma, in Mathematica one might look at Series[((1 + u)^(1/4) - 1)/u, {u, 0, 10}], say. (Mathematically, one still must worry about whether the series in fact converges in some interval about 0.) –  murray Apr 19 at 3:59
    
@murray Yes, Artes's answer points out the use of Limit and Series. I didn't think the lemma different enough to warrant pointing it out here. –  Michael E2 Apr 19 at 11:21
    
@MichaelE2 I like it i.e. geometric reasoning. –  Artes Apr 19 at 12:39
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