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Question
part (a);
calulate limit (1/x-2)^1/x-3
x->3^+
part(b)
caclulate limit (1 +3/x)^2x
x->infinity
(a) Note that A^B is the same as e^(B*ln(A)). Lets take A as 1/(x-2) and B as 1/(x-3).
)
This makes (1/(x-2))^1/(x-3) the same as e^[{1/(x-3)}ln{1/(x-2)}].
This is the same as e^[ln{1/(x-2)}/(x-3)].
Next, note that ln(1/C) = -ln(C), so this is the same as e^[-ln(x-2)/(x-3)].
Since x-3 goes to 0 as x->3 and ln(x-2)->0 as x->3, we can apply L'Hospital's Rule.
That says to calculate the limt of f(x)/g(x) as x->3, if f and g both go to 0,
we can calculate the limit on f'/g'.
Since f(x) =ln(x-2) and g(x) = x-3, f'(x) = 1/(x-2) and g'(x) = 1.
As x=3, f'(x) = 1 and g'(x) = 1, so the limit as x->3 is 1/1 = 1.
Now looking back at the problem, we have e^(-f/g), so the limit is e^-1, which is 1/e.
Checking this out in Excel, that is what I got, as can be seen in my attachment.
(b) As was mentioned in (a), this is the same as e^(2x*ln(1 + 3/x)).
Letting f(x) = 1/(2x) and g(x) = ln(1 + 3/x), we have e^(g(x)/f(x)).
It can be seen as x->∞, both f(x) and g(x) go to 0.
It can be seen that g'(x) = (-3/x²)/(1 + 3/x) and f'(x) = -1/(2x²).
Looking at the lim(x->∞)(g'/f') is lim(x->∞)[ {(-3/x²)/(1 + 3/x)}/{-1/(2x²)} ].
Multiplying by x²/x² gives lim(x->∞)[{(-3)/(1 + 3/x)}/{-1/2}].
Simplifying the fraction means inverting -1/2 and multiplying it by {(-3/(1 + 3/x)}.
Since two - makes a +, we get 6/(1 + 3/x).
Clearly, as x->∞, this becomes 3/x -> 0, so 6/(1 + 3/x) -> 6/1 -> 6.
I aslo did this in Excel, and that was what I got, as can also be seen.
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