Find the limit of this

Question

How would I compute the limit of a function such as this? $$\lim_{m\to \infty}\frac{m}{((x+m)(g-1)-m(g-2))(1+x+m)-m^2}$$

I'm not sure what to divide by in this case as I would just get the limit of the numerator to be either $0$ or $\infty$. The solution should be :

$$\lim_{m\to \infty} f(m) = \frac{1}{1+gx}$$

I'm thinking to get to this, $f(m)$ should be simplified further.

Yes, please simplify the denominator of $f(m)$. – Sarvesh Ravichandran Iyer Mar 14 '18 at 23:24 — Sarvesh Ravichandran Iyer, Mar 14 '18 at 23:24

score 1 · Answer 1 · answered Mar 28 '18 at 22:58

Compute the derivative of the denominator, call it $F(m)$: \begin{align} F(m) &=(g-1-g+2)(1+x+m)+(x+m)(g-1)-m(g-2)-2m\\ &=1+x+x(g-1)+m(1+g-1-g+2-2)\\ &=1+xg \end{align} Now it's just l'Hôpital. Or, if you prefer, $F(m)=(1+xg)m+A$, for some constant term $A$, precisely $$ A=F(0)=x(g-1)(1+x) $$ but it's unimportant.

score 0 · Accepted Answer · answered Mar 14 '18 at 23:38

0

Note that

$$\lim_{m\to \infty}\frac{m}{g m x + g x^2 + g x + m - x^2 - x}=\\=\lim_{m\to \infty}\frac{1}{g x + (g x^2)/m + (g x)/m + 1 - x^2/m - x/m}=\frac1{gx+1}$$

answered Mar 14 '18 at 23:38

user

154,566

1

Thanks, I was just about to comment my answer - simplifying the denominator made it much more easy than it was meant to be by just dividing through by m... – Btzzzz Mar 14 '18 at 23:40
You are welcome! Bye – user Mar 14 '18 at 23:43

Find the limit of this

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