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Find 
lim_(x rarr-4)(x^(2)-15)/(x^(2)-16)
Choose 1 answer:
(A) 
-(1)/(8)
(B) 
(1)/(8)
(c) 
(19)/(20)
(D) The limit doesn't exist

Find limx4x215x216 \lim _{x \rightarrow-4} \frac{x^{2}-15}{x^{2}-16} \newlineChoose 11 answer:\newline(A) 18 -\frac{1}{8} \newline(B) 18 \frac{1}{8} \newline(c) 1920 \frac{19}{20} \newline(D) The limit doesn't exist

Full solution

Q. Find limx4x215x216 \lim _{x \rightarrow-4} \frac{x^{2}-15}{x^{2}-16} \newlineChoose 11 answer:\newline(A) 18 -\frac{1}{8} \newline(B) 18 \frac{1}{8} \newline(c) 1920 \frac{19}{20} \newline(D) The limit doesn't exist
  1. Identify the limit form: Identify the form of the limit.\newlineWe need to find the limit of the function as xx approaches 4-4. Let's substitute x=4x = -4 into the function to see if we can directly evaluate the limit.\newlinef(x)=x215x216f(x) = \frac{x^2 - 15}{x^2 - 16}\newlinef(4)=(4)215(4)216f(-4) = \frac{(-4)^2 - 15}{(-4)^2 - 16}\newlinef(4)=16151616f(-4) = \frac{16 - 15}{16 - 16}\newlinef(4)=10f(-4) = \frac{1}{0}\newlineThis results in a division by zero, which is undefined. Therefore, we cannot directly evaluate the limit by substitution.
  2. Substitute x=4x = -4: Factor the denominator and simplify if possible.\newlineThe denominator x216x^2 - 16 is a difference of squares and can be factored as (x+4)(x4)(x + 4)(x - 4).\newlinef(x)=x215(x+4)(x4)f(x) = \frac{x^2 - 15}{(x + 4)(x - 4)}\newlineSince we are looking for the limit as xx approaches 4-4, we can check if the numerator has a factor of (x+4)(x + 4) that would cancel out the denominator's (x+4)(x + 4).\newlineHowever, the numerator x215x^2 - 15 does not factor with x+4x + 4 as a factor, so we cannot simplify the function by canceling out terms.
  3. Division by zero: Evaluate the limit using algebraic manipulation.\newlineSince we cannot simplify the function by canceling out terms, we need to use another method to evaluate the limit. We can use the fact that the function is continuous everywhere except at x=4x = 4 and x=4x = -4 (where the denominator is zero).\newlineFor values of xx close to but not equal to 4-4, the function is defined and continuous. Therefore, we can evaluate the limit by approaching 4-4 from either side.\newlineAs xx approaches 4-4, the numerator approaches (4)215=1615=1(-4)^2 - 15 = 16 - 15 = 1, and the denominator approaches (4)216=1616=0(-4)^2 - 16 = 16 - 16 = 0, but since we are taking a limit, we are not actually dividing by zero.\newlineThe limit of the numerator as xx approaches 4-4 is x=4x = -411, and the limit of the denominator as xx approaches 4-4 is a very small number close to zero but not equal to zero. Therefore, the limit of the function as xx approaches 4-4 is a very large number.\newlineSince the denominator is x=4x = -466, and we are approaching 4-4, the sign of the denominator is positive (because x=4x = -488). Therefore, the limit is positive.\newlineThe limit of the function as xx approaches 4-4 is positive infinity.

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