Find the limits: (i) $\lim_{x o 1}\left[\dfrac{x^2 + 1}{x + 100} ight]$ (ii) $\lim_{x o 2}\left[\dfrac{x^3 - 4x^2 + 4x}{x^2 - 4} ight]$ (iii) $\lim_{x o 2}\left[\dfrac{x^2 - 4}{x^3 - 4x^2 + 4x} ight]$ (iv) $\lim_{x o 2}\left[\dfrac{x^3 - 2x^2}{x^2 - 5x + 6} ight]$ (v) $\l

Question

Find the limits: (i) $\lim_{x	o 1}\left[\dfrac{x^2 + 1}{x + 100}ight]$ (ii) $\lim_{x	o 2}\left[\dfrac{x^3 - 4x^2 + 4x}{x^2 - 4}ight]$ (iii) $\lim_{x	o 2}\left[\dfrac{x^2 - 4}{x^3 - 4x^2 + 4x}ight]$ (iv) $\lim_{x	o 2}\left[\dfrac{x^3 - 2x^2}{x^2 - 5x + 6}ight]$ (v) $\lim_{x	o 1}\left[\dfrac{x - 2}{x^2 - x} - \dfrac{1}{x^3 - 3x^2 + 2x}ight]$

Accepted Answer

> [!TLDR]
> For a rational function, if the denominator is non-zero at the point, substitute directly; if substitution gives $\frac{0}{0}$, factor and cancel the common factor, then re-evaluate. The answers are: (i) $\frac{2}{101}$, 
> (ii) $0$, 
> (iii) does not exist (diverges to $\pm\infty$), 
> (iv) $-4$, 
> (v) $2$.

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### Concept and Intuition

For a rational function $\frac{P(x)}{Q(x)}$ as $x 	o c$: if $Q(c) 
eq 0$, 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) $\lim_{x	o 1}\left[\dfrac{x^2 + 1}{x + 100}ight]$

1. **Check for direct substitution.** At $x = 1$ the denominator is $1 + 100 = 101 
eq 0$, so the function is continuous here.
2. **Evaluate directly.**
   $$\frac{1^2 + 1}{1 + 100} = \frac{2}{101}.$$

> [!TIP]
> Whenever the denominator is non-zero at the limit point, you're done — just substitute.

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### (ii) $\lim_{x	o 2}\left[\dfrac{x^3 - 4x^2 + 4x}{x^2 - 4}ight]$

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 $(x-2)$:**
   $$\frac{x(x - 2)^2}{(x - 2)(x + 2)} = \frac{x(x - 2)}{x + 2}, \quad x 
eq 2.$$
5. **Re-evaluate at $x = 2$:**
   $$\frac{2(2 - 2)}{2 + 2} = \frac{0}{4} = 0.$$

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### (iii) $\lim_{x	o 2}\left[\dfrac{x^2 - 4}{x^3 - 4x^2 + 4x}ight]$

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 $(x-2)$:**
   $$\frac{(x - 2)(x + 2)}{x(x - 2)^2} = \frac{x + 2}{x(x - 2)}, \qu

Q.Find the limits:

Concept understanding — Algebra of Limits

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

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

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

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

Concept and Intuition

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

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

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

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

Q.Find the limits:

Concept and Intuition

(i) lim⁡x→1[x2+1x+100]\lim_{x\to 1}\left[\dfrac{x^2 + 1}{x + 100}\right]limx→1​[x+100x2+1​]

(ii) lim⁡x→2[x3−4x2+4xx2−4]\lim_{x\to 2}\left[\dfrac{x^3 - 4x^2 + 4x}{x^2 - 4}\right]limx→2​[x2−4x3−4x2+4x​]

(iii) lim⁡x→2[x2−4x3−4x2+4x]\lim_{x\to 2}\left[\dfrac{x^2 - 4}{x^3 - 4x^2 + 4x}\right]limx→2​[x3−4x2+4xx2−4​]

(iv) lim⁡x→2[x3−2x2x2−5x+6]\lim_{x\to 2}\left[\dfrac{x^3 - 2x^2}{x^2 - 5x + 6}\right]limx→2​[x2−5x+6x3−2x2​]

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

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

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

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

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