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 $(3x+7)^{\frac{1}{3x+7} \cdot (3x+7) \cdot\frac{1}{x^2-4}}$ ?
Let assume $$L=\lim_{x \rightarrow -2}(3x+7)^{\frac1{x^2-4}}$$

We can see $$\ln L=\lim_{x \rightarrow -2}\frac{\ln(3x+7)}{{x^2-4}}$$

$$\frac{\ln(3x+7)}{{x^2-4}}=\frac{\ln(1+3(x+2))}{{(x-2)(x+2)}}$$

If we use Taylor expansion at x+2, we get 

$$\frac{\ln(3x+7)}{{x^2-4}}=\frac{\ln(1+3(x+2))}{{(x-2)(x+2)}}=\frac{3(x+2))}{{(x-2)(x+2)}}-\frac{(3(x+2)))^2}{2{(x-2)(x+2)}}+O((x+2)^2) \rightarrow -\frac3{4}$$

This gives $L=e^{-\frac34}$

Or if not using Taylor expansion, let's use substitution $x=\frac13y-2$

$$\frac{\ln(3x+7)}{{x^2-4}}=\frac{\ln(1+y)}{{y}}\frac1{(\frac19y-\frac43)}$$

Keep this in mind, $$e=\lim_{h\to 0} \left(1+h\right)^{1/h}$$

The limit $L=e^{-\frac34}$
Taking log we have: $\dfrac{\log(3x+7)}{x^2-4}$, and you can use L'hospitale rule for this.