Lim as x approaches 2 of (x^2-kx+4)/(x^3-8)
To find the limit as x approaches 2 of the given expression, we can apply direct substitution first. However, if plugging in 2 results in an undefined expression (such as 0/0 or ∞/∞), we need to use additional techniques.
Let's first evaluate the expression by directly plugging in x = 2:
[(2^2 - k * 2 + 4) / (2^3 - 8)]
Simplifying further:
[(4 - 2k + 4) / (8 - 8)]
= (8 - 2k) / 0
We can observe that we have an undefined expression since we have division by zero. In this case, we need to use algebraic manipulation to simplify and find the limit.
To simplify the expression, we can factorize the numerator and denominator:
[(x^2 - kx + 4) / (x^3 - 8)]
Factoring numerator:
[(x - 2)(x - 2 + 2k) / (x^3 - 8)]
= [(x - 2)(x + 2k - 2) / (x^3 - 8)]
Now, since we want to find the limit as x approaches 2, we can rewrite the expression with x as 2:
[(2 - 2)(2 + 2k - 2) / (2^3 - 8)]
= (0)(2k) / (8 - 8)
= 0 / 0
Again, we encounter an undefined expression. To proceed, we can apply L'Hôpital's rule, which states that if the limit of f(x)/g(x) as x approaches a results in 0/0 or ∞/∞, then the limit of [f'(x)/g'(x)] as x approaches a is equal to the original limit.
To apply L'Hôpital's rule, we differentiate the numerator and denominator separately with respect to x, and then take the limit again:
Differentiating numerator:
[(2 - 2 + 2k)]
= (2k)
Differentiating denominator:
[3x^2]
= (3 * 2^2)
= 12
Now, we can find the limit again with the new expression:
Lim as x approaches 2 of (2k / 12)
= 2k / 12
= k / 6
Therefore, the limit as x approaches 2 of (x^2 - kx + 4) / (x^3 - 8) is k/6.