Q.Find the limits:

(i) $\lim_{x\to 1}\dfrac{x^2 + 1}{x + 100}$

Denominator $1+100=101\neq0$, so substitute directly: $\dfrac{1^2+1}{101}=\dfrac{2}{101}$.

(ii) $\lim_{x\to 2}\dfrac{x^3 - 4x^2 + 4x}{x^2 - 4}$

$\frac{0}{0}$: numerator $=x(x-2)^2$, denominator $=(x-2)(x+2)$. Cancel $(x-2)$: $\dfrac{x(x-2)}{x+2}\to\dfrac{2\cdot0}{4}=0$.

(iii) $\lim_{x\to 2}\dfrac{x^2 - 4}{x^3 - 4x^2 + 4x}$

$\frac{0}{0}$: $\dfrac{(x-2)(x+2)}{x(x-2)^2}=\dfrac{x+2}{x(x-2)}$. At $x=2$ this is $\dfrac{4}{0}$ — unbounded ($+\infty$ from the right, $-\infty$ from the left), so the limit does not exist.

(iv) $\lim_{x\to 2}\dfrac{x^3 - 2x^2}{x^2 - 5x + 6}$

$\frac{0}{0}$: $\dfrac{x^2(x-2)}{(x-2)(x-3)}=\dfrac{x^2}{x-3}\to\dfrac{4}{-1}=-4$. …

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Concept and Intuition

For a rational function $\frac{P(x)}{Q(x)}$ as $x \to c$: if $Q(c) \neq 0$, the limit is simply $\frac{P(c)}{Q(c)}$. If both $P(c)=0$ and $Q(c)=0$, we get the indeterminate form $\frac{0}{0}$ — this signals a common factor $(x-c)$ in top and bottom that we must cancel before re-evaluating. If only the denominator is zero, the function blows up and the limit does not exist (it tends to $\pm\infty$).

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(i) $\lim_{x\to 1}\left[\dfrac{x^2 + 1}{x + 100}\right]$

1. Check for direct substitution. At $x = 1$ the denominator is $1 + 100 = 101 \neq 0$, so the function is continuous here.
2. Evaluate directly.

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(ii) $\lim_{x\to 2}\left[\dfrac{x^3 - 4x^2 + 4x}{x^2 - 4}\right]$

1. Try substitution. Denominator: $2^2 - 4 = 0$; numerator: $8 - 16 + 8 = 0$. We have $\frac{0}{0}$.
2. Factor the numerator: $x^3 - 4x^2 + 4x = x(x^2 - 4x + 4) = x(x - 2)^2.$
3. Factor the denominator: $x^2 - 4 = (x - 2)(x + 2).$
4. Cancel the common factor $(x-2)$:

5. Re-evaluate at $x = 2$:

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(iii) $\lim_{x\to 2}\left[\dfrac{x^2 - 4}{x^3 - 4x^2 + 4x}\right]$

1. Try substitution. Numerator: $4 - 4 = 0$; denominator: $8 - 16 + 8 = 0$. Again $\frac{0}{0}$.
2. Factor both: numerator $x^2 - 4 = (x-2)(x+2)$; denominator $x(x-2)^2$.
3. Cancel $(x-2)$:

4. Re-evaluate at $x = 2$:

A non-zero numerator over zero means the function is unbounded. As $x \to 2^{+}$, $x(x-2) > 0$ so the fraction $\to +\infty$; as $x \to 2^{-}$, $x(x-2) < 0$ so it $\to -\infty$. The two one-sided limits differ, so the limit does not exist.

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(iv) $\lim_{x\to 2}\left[\dfrac{x^3 - 2x^2}{x^2 - 5x + 6}\right]$

1. Substitution check. Denominator: $4 - 10 + 6 = 0$; numerator: $8 - 8 = 0$. $\frac{0}{0}$.
2. Factor numerator: $x^3 - 2x^2 = x^2(x - 2)$.
3. Factor denominator: $x^2 - 5x + 6 = (x - 2)(x - 3)$.
4. Cancel $(x-2)$: …