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

where dā†’āˆž and R is constant.

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

Edit:

Thank you all for answer.

But now, I have new doubt

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

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  • $\begingroup$ 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? $\endgroup$ – 5xum Mar 16 '16 at 10:59
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    $\begingroup$ $$\frac{d-R}{d-2R}=\frac{1-\frac Rd}{1-\frac{2R}d}$$ $\endgroup$ – Cm7F7Bb Mar 16 '16 at 11:00
  • $\begingroup$ alternatively use L'Hopital $\endgroup$ – T'x Mar 16 '16 at 11:03
  • $\begingroup$ @Cm7F7Bb Thanks. $\endgroup$ – Anubhav Goel Mar 16 '16 at 11:11
  • $\begingroup$ 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! $\endgroup$ – christina_g Mar 16 '16 at 13:40
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Notice that the numerator and denominator of your function are polynomials. In other words f is a rational function. In that case, we have:

$$ f(x) = \frac{a_nx^n+...+a_1x+a_0}{b_mx^m+...+b_1x+b_0} \Rightarrow$$ 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 your case, apply the formula above and the wanted limit equals 1.

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Usually, with limits to infinity, the best strategy is to factor out the "dominant" term. "Dominant" isn't really correct, but for the sake of the intuition, think about it as the biggest term in the expression.

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 $0$.

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  • $\begingroup$ Its OK. But how did you cancel d/d since d/d = āˆž/āˆž which may be anytthing $\endgroup$ – Anubhav Goel Mar 16 '16 at 11:30
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    $\begingroup$ @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. $\endgroup$ – rubik Mar 16 '16 at 19:20

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