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$\lim_{x-> \infty } \left (\frac{x+a }{x+b} \right ) ^{(x+b)}$
in Calculus by (7 points)
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$e^{a-b} ?$

0
It is not in Indeterminate form.then how you got $e^{a-b}?$
0
It is in 1^(infinity) form.Is it not?
0

see this and this

$\left(\dfrac{\infty}{\infty}\right)^{\infty}$

$\implies \infty^{\infty}$, So it is not a indeterminate form.

+1
In (x+a)/(x+b), divide numerator and Denominator by 'x' and then check its form.. it is 1^infinity form
+1
Yes you are right.
0
Yes ur answer is correct can you please share the solution. .
+2
$\\ \lim_{x\rightarrow \infty } \dfrac {x(1+a/x)}{x(1+b/x)}^{x+b} =1^\infty form\\ \\ \\ e^{\lim_{x\rightarrow \infty}\big(x+b\big) \bigg(\dfrac {x+a}{x+b}-1\bigg)} \\ \\ e^{\lim_{x\rightarrow \infty }\big(x+a-x-b\big)}\\ \\ e^{\lim_{x\rightarrow \infty }\big(a-b\big)} \\ \\ e^{a-b}$

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