Math / Calculus

Question11) if $\lim _{x \rightarrow 4} \frac{3 x-12}{x^{2}-m x+4}$ A) 3

Studdy Solution

STEP 1

1. We need to find the value of $m$ such that the limit as $x$ approaches 4 of $\frac{3x-12}{x^2 - mx + 4}$ exists.
2. The numerator $3x - 12$ can be factored as $3(x - 4)$ .
3. For the limit to exist and not be infinite, the denominator $x^2 - mx + 4$ must also have a factor of $(x - 4)$ .

STEP 2

1. Factor both the numerator and denominator.
2. Simplify the expression by canceling common factors.
3. Evaluate the limit of the simplified expression as $x$ approaches 4.
4. Determine the value of $m$ that satisfies the limit condition.

STEP 3

Factor the numerator $3x - 12$ .
$3x - 12 = 3(x - 4)$

STEP 4

Assume the denominator $x^2 - mx + 4$ has a factor of $(x - 4)$ . Therefore, the denominator should be expressible as $(x - 4)(x - k)$ for some $k$ .
$x^2 - mx + 4 = (x - 4)(x - k)$

STEP 5

Expand the right-hand side of the assumed factorization of the denominator.
$(x - 4)(x - k) = x^2 - kx - 4x + 4k = x^2 - (k+4)x + 4$

STEP 6

Compare the expanded form with the original denominator to find the value of $m$ .
$x^2 - mx + 4 = x^2 - (k+4)x + 4$
Thus, we identify:
$m = k + 4$

STEP 7

Simplify the expression by canceling the common factor $(x - 4)$ in the numerator and denominator.
$\frac{3(x - 4)}{(x - 4)(x - k)} = \frac{3}{x - k}$

STEP 8

Evaluate the limit of the simplified expression as $x$ approaches 4.
$\lim_{x \to 4} \frac{3}{x - k}$
For this limit to be finite and equal to 3, the denominator $x - k$ when $x = 4$ must yield:
$\frac{3}{4 - k} = 3$

STEP 9

Solve the equation $\frac{3}{4 - k} = 3$ for $k$ .
$\frac{3}{4 - k} = 3$
Multiplying both sides by $4 - k$ :
$3 = 3(4 - k)$ $3 = 12 - 3k$ $3k = 12 - 3$ $3k = 9$ $k = 3$

STEP 10

Using $m = k + 4$ :
$m = 3 + 4$ $m = 7$
Therefore, the value of $m$ is $7$ .

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