What is the limit as x approaches 0 of #(1+2x)^cscx#?

Answer 1
The answer is #e^2#.

The reasoning is not that simple. Firstly, you must use trick: a = e^ln(a).

Therefore, #(1+2x)^(1/sinx) = e^u#, where #u=ln((1+2x)^(1/sinx)) = ln(1+2x)/sinx#
Therefore, as #e^x# is continuous function, we may move limit: #lim_(x->0) e^u = e^ (lim_(x->0)u)#
Let us calculate limit of #u# as x approaches 0. Without any theorem, calculations would be hard. Therefore, we use de l'Hospital theorem as the limit is of type #0/0#. #lim_(x->0) f(x)/g(x) = lim_(x->0)((f'(x))/(g'(x)))# Therefore, #lim_(x->0) ln(1+2x)/sinx = 2/(2x+1)/cos(x) = 2/((2x+1)cosx) = 2#
And then, if we return to the original limit #e^ (lim_(x->0)u)# and insert 2, we get the result of #e^2#,
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Answer 2

The limit as x approaches 0 of (1+2x)^cscx is 1.

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Answer 3

The limit as x approaches 0 of (1+2x)^cscx is 1.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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