how to calculate this limit? calculus I.

Question

$$\lim_{x\to 4} \frac{\sqrt{x^2 -3x} -2}{2^x-2^4}$$

I cannot use l'hospitale rule or any other,just with modificating the limit,it's for calculus I.

Answer 1

$\lim_{x\to 4} \frac{\sqrt{x^2 -3x} -2}{2^x-2^4} =\lim_{x\to 4} \frac{\sqrt{x^2 -3x} -2}{2^x-2^4} \cdot \frac{\sqrt{x^2 -3x} +2}{\sqrt{x^2 -3x} +2}=\lim_{x\to 4} \frac{x^2 -3x -4}{(2^x-2^4)(\sqrt{x^2 -3x} +2)}=\lim_{x\to 4} \frac{(x-4)(x+1)}{(2^x-2^4)(\sqrt{x^2 -3x} +2)}=\lim_{x\to 4} \frac{x-4}{2^x-2^4} \frac{x+1}{\sqrt{x^2 -3x} +2}=\frac{1}{(2^x)'|_{x=4}} \cdot \lim_{x \rightarrow 4}\frac{x+1}{\sqrt{x^2 -3x} +2} \\ \text{ You said you are given that } \\ \lim_{u \rightarrow 0} \frac{2^u-1}{u}=\ln(2) \text{ Sub in } u=x-4 \\ \lim_{x \rightarrow 4}\frac{2^{x-4}-1}{x-4}=\ln(2) \text{ multiply both sides by } 2^4 \\ \lim_{x \rightarrow 4} \frac{2^x-2^4}{x-4}=2^4 \ln(2)\\ $