Limit of an exponent function

Question

maybe you can help me solve this limit? Thanks!

$$ \lim_ {x \to \infty} \left( 1+ \frac {1}{x^2}\right)^{3x-4}$$

Answer 1

Hint: $$\left(1+\frac1{x^2}\right)^{3x-4}=\left(\left(1+\frac1{x^2}\right)^{x^2}\right)^{\frac{3x-4}{x^2}}.$$

Answer 2

Hint Let $$ f(x)=\left( 1+ \frac {1}{x^2}\right)^{3x-4} $$

Take the logarithm to get that $$ \lim_{x\to\infty}\log f(x)=\lim_{x\to\infty}\frac{3x-4}{x^2}\times\lim_{x\to\infty}\frac{\log(1+x^{-2})-\log 1}{x^{-2}} $$ The first limit should be easity to compute, the second follows from the definition of the derivative.