Limits - a different approach $\lim_{x \to\infty }(\frac{x^3+4x^2+3x+5}{x^2+2x+3}+lx+m)=10$.
How do I calculate the value of l and m? 
My try: I know questions having limit tending to infinity can be solved by dividing the numerator and denominator by greatest power of $x$.But it got me nowhere in this question. Any help appreciated.
 A: $$\lim\limits_{x \to \infty}(\frac{x^3+4x^2+3x+5}{x^2+2x+3}+lx+m)=10$$
$$\lim\limits_{x \to \infty}(\frac{x^3+4x^2+3x+5+lx^3+2lx^2+3lx+mx^2+2mx+3m}{x^2+2x+3})=10$$
$$\lim\limits_{x \to \infty}(\frac{(1+l)x^3+(4+2l+m)x^2+(3+l+2m)x+(5+3m)}{x^2+2x+3})=10$$
Note that as the limit exists, the coefficient of $x^3$ has to be $0$, so we get $l=-1$
And also the ratio of the leading coefficient ($x^2$)$=10$, so we get $4+2l+m=10$ and thus $m=8$
A: Using the long division, you have $$\frac{x^3+4 x^2+3 x+5}{x^2+2 x+3}=x+2-\frac{4}{x}+\cdots$$ So $$\lim_{x \to\infty }\left(\frac{x^3+4x^2+3x+5}{x^2+2x+3}+lx+m\right)=\lim_{x \to\infty }(x+2+lx+m)=10$$ from which you can conclude that $l=-1$ and $m=8$.
A: If this limit is $10$, then the limit when you divide by $x$ is $0$: so
$$
\lim_{x \to\infty }\left(\frac{x^3+4x^2+3x+5}{x(x^2+2x+3)}+l+\frac{m}{x}\right)=0
$$
Therefore $1+l=0$; now
$$
\lim_{x \to\infty }\left(\frac{x^3+4x^2+3x+5}{x^2+2x+3}-x+m\right)=
\lim_{x \to\infty }\left(\frac{2x^2+5}{x^2+2x+3}+m\right)=2+m
$$
A: Indeed, we have
$$\lim_{x\to\infty }\left(\frac{x^3+4x^2+3x+5}{x^2+2x+3}-[-lx+10-m]\right)=0$$.
$$-l=\lim_{x\to\infty }\frac{\frac{x^3+4x^2+3x+5}{x^2+2x+3}}{x}=1$$
thus
$l=-1$
and 
$$10-m=\lim_{x\to\infty }\left(\frac{x^3+4x^2+3x+5}{x^2+2x+3}-x\right)=2$$
therefore $m=8$
A: $$\lim_{x \to\infty }\left(\dfrac{x^3+4x^2+3x+5}{x^2+2x+3}+\ell x + m\right)=10$$
$$\lim_{x \to\infty }\left(\dfrac{x^3+4x^2+3x+5}{x^2+2x+3}+\ell x\right)=10-m$$
$$\lim_{x \to\infty }\left(x + \dfrac{2x^2+5}{x^2+2x+3}+\ell x\right)=10-m$$
$$\lim_{x \to\infty }(x +\ell x)=0 \implies \ell = -1$$
$$\lim_{x \to\infty }\left(\dfrac{2x^2+5}{x^2+2x+3}\right)=10-m
\implies 2=10-m \implies m=8$$
A: $\dfrac{x^3+4x^2+3x+5}{x^2+2x+3} \approx_{\infty} x\implies \lim_{x \to \infty} (x+lx+m) = 10\implies l = -1, m = 10.$
