Final Answer
$\frac{16}{7}$
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Step-by-step Solution
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1
If we directly evaluate the limit $\lim_{x\to 4}\left(\frac{3x^2-8x-16}{2x^2-9x+4}\right)$ as $x$ tends to $4$, we can see that it gives us an indeterminate form
$\frac{0}{0}$
2
We can solve this limit by applying L'H么pital's rule, which consists of calculating the derivative of both the numerator and the denominator separately
$\lim_{x\to 4}\left(\frac{\frac{d}{dx}\left(3x^2-8x-16\right)}{\frac{d}{dx}\left(2x^2-9x+4\right)}\right)$
3
After deriving both the numerator and denominator, the limit results in
$\lim_{x\to4}\left(\frac{6x-8}{4x-9}\right)$
4
Evaluate the limit $\lim_{x\to4}\left(\frac{6x-8}{4x-9}\right)$ by replacing all occurrences of $x$ by $4$
$\frac{6\cdot 4-8}{4\cdot 4-9}$
5
Multiply $4$ times $4$
$\frac{6\cdot 4-8}{16-9}$
6
Subtract the values $16$ and $-9$
$\frac{6\cdot 4-8}{7}$
7
Multiply $6$ times $4$
$\frac{24-8}{7}$
8
Subtract the values $24$ and $-8$
$\frac{16}{7}$
9
Divide $16$ by $7$
$\frac{16}{7}$
Final Answer
$\frac{16}{7}$
Exact Numeric Answer
$2.2857$