<s>[INST] Find $\lim\limits_{x\to -2} (3x+7)^{\frac{1}{x^2-4}}$ 
$$\lim_{x\to -2} (3x+7)^{\frac{1}{x^2-4}}$$

$$y=\lim_{x\to -2} (3x+7)^{\frac{1}{x^2-4}}$$
$$\ln y=\ln \left(\lim_{x\to -2} (3x+7)^{\frac{1}{x^2-4}} \right)$$
$$\ln y=\left(\lim_{x\to -2} \frac{\ln(3x+7)}{x^2-4}\right)$$
Applying L'hopital 
$$\ln y= \left(\lim_{x\to -2} \frac{\frac{3}{3x+7}}{2x}\right)$$
$$\ln y=-\frac{3}{4}$$
$$y=e^{-\frac{3}{4}}$$
Is there a way to solve it using something like $(3x+7)^{\frac{1}{3x+7} \cdot (3x+7) \cdot\frac{1}{x^2-4}}$ ?
 [/INST] Taking log we have: $\dfrac{\log(3x+7)}{x^2-4}$, and you can use L'hospitale rule for this. 
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