Limit of $\log_{4n^2-2}(1-(5n^2-7n+3)/(8n^2-8n+2))+2$ with $n\to\infty$

Question

Can someone please help me find the limit of this equation? $$\lim_{n\to\infty}\log_{4n^2-2}\left(1-\frac{5n^2-7n+3}{8n^2-8n+2}\right)+2$$

Answer 1

Two ways to answer.

1) Take the function $y=\log_x(a)$ with $0\lt a\lt 1$. It is easy to verify that the limit of $y$ when $x$ tends to $\infty$ is equal to $0$. Now one has $$1-\frac{5n^2-7n+3}{8n^2-8n+2}\le 0.542\lt 1$$ Thus the asked limit is equal to $0+2=\color{red}2$

2) One has $$f(n)-2=\log_{4n^2-2}\left(1-\frac{5n^2-7n+3}{8n^2-8n+2}\right)\iff(4n^2-2)^{f(n)-2}=\left(1-\frac{5n^2-7n+3}{8n^2-8n+2}\right)$$ it follows, going to the limit, the form $$\infty^{f(n)-2}=1-\frac 58=\frac 38$$ This is compatible only with $f(n)-2=0$ hence the limit of $f(n)$ when $n$ tends to $\infty$ is equal to $\color{red}2$

Answer 2

Note that $$ \lim_{n\rightarrow\infty} (4n^2 -2)^{\log_{4n^2-2}\left(1-\frac{5n^2-7n+3}{8n^2-8n+2}\right)} = \lim_{n\rightarrow\infty} \left(1-\frac{5n^2-7n+3}{8n^2-8n+2}\right) = 1-\frac{5}{8} = \frac{3}{8}. $$ Since $4n^2 - 2 \rightarrow\infty$ we must have that the logarithm goes to zero.

Answer 3

Well, you can start by writing, for $n>0$,

$\log_{4n^2-2}\bigg(1-\frac{5n^2-7n+3}{8n^2-8n+2}\bigg) + 2=2+\dfrac{\log\bigg(1-\frac{5n^2-7n+3}{8n^2-8n+2}\bigg)}{\log\bigg(4n^2-2\bigg)}$

I think you can take it from here....

Answer 4

$\lim_{n\rightarrow\infty} = \log_{4n^2-2}\bigg(1-\frac{5n^2-7n+3}{8n^2-8n+2}\bigg) + 2 =\lim_{n\rightarrow\infty} = \log_{4n^2-2}\bigg(\frac{3n^2-n-1}{8n^2-8n+2}\bigg) + 2$=$\log_{4n^2-2} (3n^2-n-1) - \log_{4n^2-2} (8n^2-8n+2) +2$ =$\frac {log_{}(3n^2-n-1)} {log (4n^2-2)}-\frac {log_{}(8n^2-8n+2)} {log (4n^2-2)}+2$

can we use L'hopital?