Q.Evaluate:

Both limits are $\dfrac{0}{0}$, so simplify before substituting.

(i) Use $x^n-1=(x-1)(x^{n-1}+\cdots+x+1)$ to factor and cancel $(x-1)$:

At $x=1$ the top has $15$ terms and the bottom has $10$, each equal to $1$, so the value is $\dfrac{15}{10}=\dfrac{3}{2}$.

(ii) Rationalise with the conjugate $\sqrt{1+x}+1$: …

Why these techniques work

When direct substitution of the limiting value gives $\dfrac{0}{0}$, the numerator and denominator must share a common factor that is making both vanish. Our job is to expose that factor and cancel it, after which ordinary substitution works.

(i) $\lim_{x\to 1}\dfrac{x^{15} - 1}{x^{10} - 1}$

Step 1 — Check direct substitution. At $x=1$: numerator $=1-1=0$, denominator $=1-1=0$. This is $\dfrac{0}{0}$, so we simplify first.

Step 2 — Factor using the difference-of-powers identity. For any positive integer $n$,

Applying it to both:

Step 3 — Cancel the common factor. For $x\neq 1$ we may cancel $(x-1)$:

Step 4 — Substitute $x=1$. The numerator has $15$ terms, each equal to $1$, so it becomes $15$; the denominator has $10$ terms, each equal to $1$, so it becomes $10$:

(ii) $\lim_{x\to 0}\dfrac{\sqrt{1 + x} - 1}{x}$ …