# How to solve infinity by infinity here? [closed]

$f(d) = \frac{d - R}{d - 2R}$

where dāā and R is constant.

I tried to use method used for 0/0 form. But I failed and got 0 answer.

Edit:

But now, I have new doubt

how, could we get d/d = 1. Since, it may be anything.

• What is "method used for 0/0 form"? How did you fail? Also, what is $x$? Why is there only $x$ on one side of the equation? What precisely are you trying to calculate? The limit? – 5xum Mar 16 '16 at 10:59
• $$\frac{d-R}{d-2R}=\frac{1-\frac Rd}{1-\frac{2R}d}$$ – Cm7F7Bb Mar 16 '16 at 11:00
• alternatively use L'Hopital – T'x Mar 16 '16 at 11:03
• @Cm7F7Bb Thanks. – Anubhav Goel Mar 16 '16 at 11:11
• When you have a limit of the form $0/0$ or $\infty / \infty$ the result is essentially "how many times faster than the numerator does the denominator converge to 0 (or $\infty$) ". In your case since both the numinator and the denominator are polynomials of degree 1 with the same coefficient, the rate of conergence of the one is equal to the rate of convergence of the other. In polynomials, the rate of convergence depends ONLY on the coefficient of the term of the largest degree, and the degree itself! – christina_g Mar 16 '16 at 13:40

$$f(x) = \frac{a_nx^n+...+a_1x+a_0}{b_mx^m+...+b_1x+b_0} \Rightarrow$$ if $m=n$ $$\lim_{x \rightarrow \infty} f(x) = \frac {a_n}{b_m}$$ if $m>n$ $$\lim_{x \rightarrow \infty} f(x) = 0$$ if $m<n$ $$\lim_{x \rightarrow \infty} f(x) = \infty$$
In this case, we have two terms, $d$ and $R$. The second one is constant, while the first one grows to infinity. Therefore the dominant term is $d$. So $$\require{cancel}\lim_{d \to +\infty} \frac{d - R}{d - 2R} = \lim_{d \to +\infty} \frac{\cancel{d}(1 - R/d)}{\cancel{d}(1 - 2R/d)} = 1$$ since both $R/d$ and $2R/d$ tend to $0$.
• @AnubhavGoel $d$ is not $\infty$, since $d$ is a real number and $\infty \notin \mathbb R$. Note that we simplify the expression before evaluating the limit. So we are cancelling two numbers (we don't exactly know their value, but that's not a problem). The important thing is that $d \neq 0$, since we are working in a neighborhood of $+\infty$. Note that this method is the one that it's used to prove the claim in the accepted answer. – rubik Mar 16 '16 at 19:20