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Notely Maths 12
Class 12 Maths
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1.4 Q-1

Question Statement

Determine the left-hand limit (LHL), right-hand limit (RHL), and overall limit for the following functions as x→cx \to c:

  1. f(x)=2x2+xβˆ’5,,c=1f(x) = 2x^2 + x - 5, , c = 1
  2. f(x)=x2βˆ’9xβˆ’3,,c=βˆ’3f(x) = \frac{x^2 - 9}{x - 3}, , c = -3
  3. f(x)=∣xβˆ’5∣,,c=5f(x) = |x - 5|, , c = 5

Background and Explanation

To find limits, you need to evaluate how a function behaves as xx approaches a specific value (cc) from the left (xβ†’cβˆ’x \to c^-) and the right (xβ†’c+x \to c^+). If both LHL and RHL are equal, the limit exists and equals their value. For functions involving polynomials, rational expressions, or absolute values, factorization, simplification, and substitution are commonly used to evaluate limits.

Solution

1. f(x)=2x2+xβˆ’5,,c=1f(x) = 2x^2 + x - 5, , c = 1

Left-Hand Limit (LHL)

LHL=lim⁑xβ†’1βˆ’f(x)=lim⁑xβ†’1βˆ’(2x2+xβˆ’5)\text{LHL} = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (2x^2 + x - 5)

Substitute x=1x = 1:

LHL=2(1)2+1βˆ’5=2+1βˆ’5=βˆ’2\text{LHL} = 2(1)^2 + 1 - 5 = 2 + 1 - 5 = -2

Right-Hand Limit (RHL)

RHL=lim⁑xβ†’1+f(x)=lim⁑xβ†’1+(2x2+xβˆ’5)\text{RHL} = \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2x^2 + x - 5)

Substitute x=1x = 1:

RHL=2(1)2+1βˆ’5=2+1βˆ’5=βˆ’2\text{RHL} = 2(1)^2 + 1 - 5 = 2 + 1 - 5 = -2

Overall Limit

Since LHL=RHL=βˆ’2\text{LHL} = \text{RHL} = -2, the overall limit is:

lim⁑xβ†’1f(x)=βˆ’2\lim_{x \to 1} f(x) = -2

2. f(x)=x2βˆ’9xβˆ’3,,c=βˆ’3f(x) = \frac{x^2 - 9}{x - 3}, , c = -3

Left-Hand Limit (LHL)

LHL=lim⁑xβ†’βˆ’3βˆ’f(x)=lim⁑xβ†’βˆ’3βˆ’x2βˆ’9xβˆ’3\text{LHL} = \lim_{x \to -3^-} f(x) = \lim_{x \to -3^-} \frac{x^2 - 9}{x - 3}

Factorize x2βˆ’9=(xβˆ’3)(x+3)x^2 - 9 = (x - 3)(x + 3):

LHL=lim⁑xβ†’βˆ’3βˆ’(xβˆ’3)(x+3)xβˆ’3\text{LHL} = \lim_{x \to -3^-} \frac{(x - 3)(x + 3)}{x - 3}

Cancel xβˆ’3x - 3 (valid as xβ‰ 3x \neq 3):

LHL=lim⁑xβ†’βˆ’3βˆ’(x+3)\text{LHL} = \lim_{x \to -3^-} (x + 3)

Substitute x=βˆ’3x = -3:

LHL=βˆ’3+3=0\text{LHL} = -3 + 3 = 0

Right-Hand Limit (RHL)

RHL=lim⁑xβ†’βˆ’3+f(x)=lim⁑xβ†’βˆ’3+x2βˆ’9xβˆ’3\text{RHL} = \lim_{x \to -3^+} f(x) = \lim_{x \to -3^+} \frac{x^2 - 9}{x - 3}

Repeat the same steps:

RHL=lim⁑xβ†’βˆ’3+(x+3)=βˆ’3+3=0\text{RHL} = \lim_{x \to -3^+} (x + 3) = -3 + 3 = 0

Overall Limit

Since LHL=RHL=0\text{LHL} = \text{RHL} = 0, the overall limit is:

lim⁑xβ†’βˆ’3f(x)=0\lim_{x \to -3} f(x) = 0

3. f(x)=∣xβˆ’5∣,,c=5f(x) = |x - 5|, , c = 5

Left-Hand Limit (LHL)

LHL=lim⁑xβ†’5βˆ’f(x)=lim⁑xβ†’5βˆ’βˆ£xβˆ’5∣\text{LHL} = \lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} |x - 5|

As xβ†’5βˆ’x \to 5^-, ∣xβˆ’5∣=βˆ’(xβˆ’5)|x - 5| = -(x - 5) because xβˆ’5<0x - 5 < 0:

LHL=lim⁑xβ†’5βˆ’βˆ’(xβˆ’5)=βˆ’(5βˆ’5)=0\text{LHL} = \lim_{x \to 5^-} -(x - 5) = -(5 - 5) = 0

Right-Hand Limit (RHL)

RHL=lim⁑xβ†’5+f(x)=lim⁑xβ†’5+∣xβˆ’5∣\text{RHL} = \lim_{x \to 5^+} f(x) = \lim_{x \to 5^+} |x - 5|

As xβ†’5+x \to 5^+, ∣xβˆ’5∣=xβˆ’5|x - 5| = x - 5 because xβˆ’5β‰₯0x - 5 \geq 0:

RHL=lim⁑xβ†’5+(xβˆ’5)=5βˆ’5=0\text{RHL} = \lim_{x \to 5^+} (x - 5) = 5 - 5 = 0

Overall Limit

Since LHL=RHL=0\text{LHL} = \text{RHL} = 0, the overall limit is:

lim⁑xβ†’5f(x)=0\lim_{x \to 5} f(x) = 0

Key Formulas or Methods Used

  • Polynomial substitution: Direct evaluation of limits by substituting xβ†’cx \to c.
  • Factorization: Simplifying rational functions by canceling common terms.
  • Absolute value definition: Splitting the function into cases based on the sign of the expression inside ∣xβˆ’a∣|x - a|.

Summary of Steps

  1. For f(x)=2x2+xβˆ’5,,c=1f(x) = 2x^2 + x - 5, , c = 1:
    • Substitute x=1x = 1 to find LHL, RHL, and the overall limit (βˆ’2-2).
  2. For f(x)=x2βˆ’9xβˆ’3,,c=βˆ’3f(x) = \frac{x^2 - 9}{x - 3}, , c = -3:
    • Factorize numerator, cancel common terms, and substitute x=βˆ’3x = -3 (00).
  3. For f(x)=∣xβˆ’5∣,,c=5f(x) = |x - 5|, , c = 5:
    • Use the definition of absolute value for LHL and RHL (00).