Let $f(x) = \frac{x - 1}{2x^2 - 7x + 5}$. Then:

$\lim _{x \rightarrow \infty}\left[\frac{8 x^2+5 x+3}{2 x^2-7 x-5}\right]^{\frac{4 x+3}{8 x-1}} = $

Let $f(x) = \frac{\ln(x^2 + e^x)}{\ln(x^4 + e^{2x})}$. If $\lim_{x \to \infty} f(x) = l$ and $\lim_{x \to -\infty} f(x) = m$,then:

Let $f(x)=5-|x-2|$ and $g(x)=|x+1|$,$x \in R$. If $f(x)$ attains its maximum value at $\alpha$ and $g(x)$ attains its minimum value at $\beta$,then $\lim _{x \rightarrow-\alpha \beta} \frac{(x-1)(x^2-5x+6)}{(x^2-6x+8)}$ is equal to

$\mathop {\lim }\limits_{x \to \infty} \frac{2x^2 + 3x + 4}{3x^2 + 3x + 4}$ is equal to

If $0 \leq x \leq \pi / 2$,then $\lim _{x \rightarrow a} \frac{|2 \cos x-1|}{2 \cos x-1}$