Why is $0$ the largest value of $a$ satisfying $\lim_{x\to1}\left(\frac{\sin(x-1)+a-ax}{\sin(x-1)+x-1}\right)^\frac{1-x}{1-\sqrt x}=\frac14$?

Question

Why is the correct value in this limit problem?

Body: I came across this problem:

Find the largest value of the nonnegative integer $a$ for which $$\lim_{x \to 1} \left( \frac{-ax + \sin(x-1) + a}{x + \sin(x-1) - 1} \right)^{\frac{1 - x}{1 - \sqrt{x}}} = \frac{1}{4} $$

The answer key gives $a = 0$, but I thought it should be $a = 2$. Here’s how I approached it:

1. Exponent simplification:

$\frac{1 - x}{1 - \sqrt{x}} = \frac{(1 - x)(1 + \sqrt{x})}{(1 - \sqrt{x})(1 + \sqrt{x})} = \frac{(1 - x)(1 + \sqrt{x})}{1 - x} = 1 + \sqrt{x}$

So as $x \to 1$, the exponent approaches 2.

---

2. Base evaluation using L'Hôpital's Rule:

$\lim_{x \to 1} \frac{-ax + \sin(x - 1) + a}{x + \sin(x - 1) - 1}$

Differentiating numerator and denominator:

Numerator: $-a + \cos(x - 1)$

Denominator: $1 + \cos(x - 1)$

At $x = 1$, this becomes:

$\frac{-a + 1}{1 + 1} = \frac{1 - a}{2}$

---

3. Final expression:

$\left( \frac{1 - a}{2} \right)^2 = \frac{1}{4} \Rightarrow \frac{(1 - a)^2}{4} = \frac{1}{4} \Rightarrow (1 - a)^2 = 1 \Rightarrow 1 - a = \pm 1 \Rightarrow a = 0 \text{ or } 2$

---

Since both values satisfy the limit, and the question asks for the largest valid $a$, I expected the answer to be $a = 2$. Why is $a = 0$ instead?

Answer 1

$a^b$ for positive non-integer $b$ is only defined when $a\ge0$, we're evaluating around a neighborhood of $x=1$, so when $\epsilon$ is very small and $1-\epsilon<x<1,1<1+\sqrt{x}<2$, when $1<x<1+\epsilon$, $2<1+\sqrt{x}<3$, therefore the exponent is never an integer, applying this to the problem: $$\frac{-ax+\sin(x-1)+a}{x+\sin(x-1)-1}\ge0$$ $$\implies \frac{-a+\frac{\sin(x-1)}{x-1}}{1+\frac{\sin(x-1)}{x-1}}\ge0$$ But when $|x-1|<\epsilon$ for very small $\epsilon$, $\frac{\sin(x-1)}{x-1}$ is positive, therefore $$-a+\frac{\sin(x-1)}{x-1}\ge 0$$ $$\implies a\le\frac{\sin(x-1)}{x-1}<1$$ So a necessary condition for the limit to exist is when $a<1$, and $a=2$ can not be a solution to this problem. Again when $|x-1|<\epsilon$ for very small $\epsilon$, $\frac{\sin(x-1)}{x-1}$ is positive, so $a=0$ satisfy the condition.

Answer 2

In real analysis, the exponential function $f(x)=a^x$ is only defined for positive $a$. This is because it does not have a well-defined behaviour for negative $a$ as its values wildly fluctuate between purely real and unreal values. For example, when $x=\frac1{2n+1},n\in\Bbb Z$, the function is purely real but for $x=\frac1{2n},n\in\Bbb Z\setminus\{0\}$, the function is not real.

So, the domain of $a$ is the solution set of the following inequality where $x\to0$:

$$\frac{\sin x-ax}{\sin x+x}>0\implies\frac{\frac{\sin x}x-a}{\frac{\sin x}x+1}>0\implies a<\frac{\sin x}x<1$$

Hence, $a=0$ is the only solution.

Answer 3

For $x\neq 1$, the base is $\ 1+ (-1-a) \frac{x-1}{\sin(x-1)+x-1}= 1- \frac{1+a}{1+\frac{\sin(x-1)}{x-1}}$ goes to $1- \frac{1+a}{2}= \frac{1-a}{2}$, as you obtained already.

For $a=2$ you may as well have $\frac{1-a}{2}= \frac{-1}{2}= \lim_{x\to 1} g(x)$ , but your function near $1$ is not defined as the base cannot be negative and variable.

You cannot consider $e^{(1+\sqrt{x})\log(g(x))}$.

Try to think about why the function $x\mapsto x^x$ is not defined for negative $x$.

Answer 4

The case for $a=2$ was already discussed here

• Evaluate the limit:$\lim_{x\to 1}\left(\frac{2-2x+\sin(x-1)}{x-1+\sin(x-1)}\right)^{\frac{1-x}{1-\sqrt x}}$

and if we refer to the more general definition of limit, we can give a meaning to the expression considering rational exponents with odd denominator reduced to coprime factors.

Therefore answer depends upon the definition we are assuming.