Q.If $\displaystyle\lim_{x\to 1}\left[\frac{x^4-1}{x-1}\right]=\lim_{x\to a}\left[\frac{x^3-a^3}{x-a}\right]$, find all possible values of $a$.

Rationalise the numerator by multiplying by its conjugate to remove the $0/0$ indeterminate form, then substitute $x=0$.

$\lim_{x\to0}\dfrac{\sqrt{1+x}-1}{x}$

Multiply numerator and denominator by $\sqrt{1+x}+1$: …

Factor the numerator using a³−b³ = (a−b)(a²+ab+b²) to cancel the (x−1) that causes the 0/0 form.

$\lim_{x\to1}\frac{x^3-1}{x-1}$

Direct substitution gives $\dfrac{0}{0}$, an indeterminate form, so factor:

$x^3-1 = (x-1)(x^2+x+1)$

So:

$\lim_{x\to1}\frac{(x-1)(x^2+x+1)}{x-1} = \lim_{x\to1}(x^2+x+1) = 1+1+1 = 3$

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Multiplying by the conjugate turns the $0/0$ form into a limit that can be evaluated directly.

$\displaystyle\lim_{x\to 0}\dfrac{\sqrt{1+x}-1}{x}$

Multiply numerator and denominator by $\sqrt{1+x}+1$:

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Factor the numerator as a difference of squares and cancel the common factor before substituting.

$4x^2-1 = (2x-1)(2x+1)$

$\dfrac{4x^2-1}{2x-1} = \dfrac{(2x-1)(2x+1)}{2x-1} = 2x+1$ (for $x\ne \tfrac12$)

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Rationalizing the numerator turns the $0/0$ form into a limit that evaluates directly to $1/2$.

Multiply numerator and denominator by the conjugate $\sqrt{1+x}+1$:

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

…

The limit equals $5$.

Expand: $(x+1)^5 = 1+5x+10x^2+10x^3+5x^4+x^5$.

Subtracting 1 and dividing by $x$:

$\dfrac{(x+1)^5-1}{x} = 5+10x+10x^2+5x^3+x^4 \quad (x e0).$

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Multiply by the conjugate of the numerator to clear the square root.

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Split into two standard limits of the form $\lim_{x\to1}\dfrac{x^n-1}{x-1}=n$.

$\dfrac{x^{15}-1}{x^{10}-1}=\dfrac{(x^{15}-1)/(x-1)}{(x^{10}-1)/(x-1)}$. As $x\to1$, the numerator's ratio $\to15$ and the denomi …

The standard limit $\lim_{x\to a}\frac{x^n-a^n}{x-a}=na^{n-1}$; the exponent in the question is off by one, so the statement is false.

Recall the standard algebraic limit:

$\lim_{x\to a}\frac{x^n-a^n}{x-a} = na^{n-1}$

This can be verified using the factorization $x^n - a^n = (x-a)(x^{n-1}+x^{n-2}a+\cdots+a^{n-1})$, which has $n$ terms, each tending to $a^{n-1}$ as $x\to a$, giving $na^{n-1}$.

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The limit equals $12$.

Factor the numerator using $a^3-b^3=(a-b)(a^2+ab+b^2)$ with $a=x,b=2$:

$x^3-8 = (x-2)(x^2+2x+4).$

So:

$\dfrac{x^3-8}{x-2} = x^2+2x+4 \quad(x e2).$

…

Rationalising the numerator turns the 0/0 form into a limit equal to $1/2$.

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

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The limit evaluates to $3$.

Expand the numerator:

$(x+1)^3-1 = x^3+3x^2+3x+1-1 = x^3+3x^2+3x = x(x^2+3x+3).$

So

$\dfrac{(x+1)^3-1}{x} = \dfrac{x(x^2+3x+3)}{x} = x^2+3x+3\quad (x e0).$

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