Binomial expansion to find limits Watch

jameshyland29
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https://mathsorchard.weebly.com/uplo...1819/1998.docx

Concerning 1998 STEP 3 question 3...

In the given solution the expression found for Vn has c as a denominator, and I'm wondering why that doesn't pose any problems for evaluating at c=0. Is it simply because everything in the bracket disappears before the denominator by rules of precedence?

Also, is the binomial expansion absolutely necessary there? Because without it it's unclear whether the final term tends to zero or infinity?

Is there no other way to deduce that the limit as c tends to zero is the same as the value when c = 0 ? Is this fact less obvious than it sounds to me to be?

Thanks.
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RDKGames
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(Original post by jameshyland29)
https://mathsorchard.weebly.com/uplo...1819/1998.docx

Concerning 1998 STEP 3 question 3...

In the given solution the expression found for Vn has c as a denominator, and I'm wondering why that doesn't pose any problems for evaluating at c=0. Is it simply because everything in the bracket disappears before the denominator by rules of precedence?
There *is* an issue with evaluating V_n at c=0, which is why the limits are used to justify the statement V_n = V_0 - Nd.

Also, is the binomial expansion absolutely necessary there? Because without it it's unclear whether the final term tends to zero or infinity?

Is there no other way to deduce that the limit as c tends to zero is the same as the value when c = 0 ? Is this fact less obvious than it sounds to me to be?

Thanks.
You don't *need* to use the binomial expansion here, but it does make it a whole lot clearer as to what the limit must be because you are then able to cancel out the c in the denominator.

An alternative approach is to use L'Hopitals rule:

\displaystyle \lim_{c \to 0} \dfrac{d[1-(1+c)^N]}{c} = \lim_{c \to 0} \dfrac{d[0-N(1+c)^{N-1}]}{1} = -Nd.

In addition to that, since \displaystyle \lim_{c \to 0} V_0(1+c)^N = V_0 that means V_n \to V_0 - Nd.
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