x2 + 3x = 28 lim X ~4 X – 4 X2 X – 2 lim X -2

x2 + 3x = 28 lim X ~4 X – 4 X2 X – 2 lim X -2

The Correct Answer and Explanation is:

1.lim⁡x→4×2+3x−28x−4\lim_{x \to 4} \frac{x^2 + 3x – 28}{x – 4}x→4lim​x−4×2+3x−28​

Step-by-step:

First, factor the numerator:x2+3x−28=(x+7)(x−4)x^2 + 3x – 28 = (x + 7)(x – 4)x2+3x−28=(x+7)(x−4)

So the expression becomes:(x+7)(x−4)x−4\frac{(x + 7)(x – 4)}{x – 4}x−4(x+7)(x−4)​

Now cancel out the common factor (x−4)(x – 4)(x−4):(x+7)(x−4)(x−4)=x+7\frac{(x + 7)\cancel{(x – 4)}}{\cancel{(x – 4)}} = x + 7(x−4)​(x+7)(x−4)​​=x+7

Now take the limit as x→4x \to 4x→4:lim⁡x→4(x+7)=4+7=11\lim_{x \to 4} (x + 7) = 4 + 7 = 11x→4lim​(x+7)=4+7=11

Answer: 11\boxed{11}11​

2.lim⁡x→214−1x2x−2\lim_{x \to 2} \frac{\frac{1}{4} – \frac{1}{x^2}}{x – 2}x→2lim​x−241​−x21​​

Step-by-step:

Combine the numerator into one fraction:14−1×2=x2−44×2\frac{1}{4} – \frac{1}{x^2} = \frac{x^2 – 4}{4x^2}41​−x21​=4x2x2−4​

Now substitute that back into the main expression:x2−44x2x−2\frac{\frac{x^2 – 4}{4x^2}}{x – 2}x−24x2x2−4​​

Recall that x2−4=(x−2)(x+2)x^2 – 4 = (x – 2)(x + 2)x2−4=(x−2)(x+2), so we have:(x−2)(x+2)4×2(x−2)=(x−2)(x+2)4×2(x−2)=x+24×2\frac{(x – 2)(x + 2)}{4x^2(x – 2)} = \frac{\cancel{(x – 2)}(x + 2)}{4x^2\cancel{(x – 2)}} = \frac{x + 2}{4x^2}4×2(x−2)(x−2)(x+2)​=4×2(x−2)​(x−2)​(x+2)​=4x2x+2​

Now take the limit as x→2x \to 2x→2:2+24(2)2=416=14\frac{2 + 2}{4(2)^2} = \frac{4}{16} = \frac{1}{4}4(2)22+2​=164​=41​

Answer: 14\boxed{\frac{1}{4}}41​​

Explanation

These problems test our understanding of evaluating limits, especially when direct substitution gives indeterminate forms like 00\frac{0}{0}00​. When that happens, we must simplify the expression.

In the first problem, the limit is:lim⁡x→4×2+3x−28x−4\lim_{x \to 4} \frac{x^2 + 3x – 28}{x – 4}x→4lim​x−4×2+3x−28​

Substituting x=4x = 4x=4 into the numerator gives 16+12−28=016 + 12 – 28 = 016+12−28=0, and the denominator is also 0, giving an indeterminate form. To resolve this, we factor the numerator. Factoring allows cancellation of the troublesome (x−4)(x – 4)(x−4) term, simplifying the expression to x+7x + 7x+7, which is continuous at x=4x = 4x=4. Evaluating the limit becomes a matter of simple substitution, resulting in 11.

In the second problem, the limit is:lim⁡x→214−1x2x−2\lim_{x \to 2} \frac{\frac{1}{4} – \frac{1}{x^2}}{x – 2}x→2lim​x−241​−x21​​

Again, direct substitution gives 14−14=0\frac{1}{4} – \frac{1}{4} = 041​−41​=0 in the numerator and x−2=0x – 2 = 0x−2=0 in the denominator. This 00\frac{0}{0}00​ form signals the need for algebraic simplification. By combining the fractions in the numerator and factoring the resulting expression, we cancel the (x−2)(x – 2)(x−2) term. The simplified form is continuous at x=2x = 2x=2, allowing us to directly evaluate the limit and find it equals 14\frac{1}{4}41​.

In both cases, factoring and simplification help eliminate indeterminate forms, making it possible to evaluate the limit accurately.