lim_(t rarr0)(1^((1)/(sin^(2)t))+2^((1)/(sin^(2)t))+dots.+n^((1)/(sin^(2)t)))^(sin^(2)t) is equal to

Given Limit Expression: We are given the limit expression lim_(t -> 0)(1^(1/sin^2(t)) + 2^(1/sin^2(t)) + ... + n^(1/sin^2(t)))^(sin^2(t)). To solve this, we will use the fact that as t approaches 0, sin^2(t) approaches 0 as well. We will also use the property that any number raised to the power of 0 is 1.Consider Base of Exponentiation: First, let's consider the base of the exponentiation, which is 1^(1/sin^2(t)) + 2^(1/sin^2(t)) + ... + n^(1/sin^2(t)). As t approaches 0, 1/sin^2(t) approaches infinity. Therefore, each term of the form k^(1/sin^2(t)) (where k is a constant) approaches 1, because any non-zero constant raised to the power of infinity is 1.Sum Approaches n: Since each term in the sum approaches 1, the entire sum approaches n, because there are n terms in the sum and each term is approaching 1.Exponent Approaches 0: Now, we have the expression (n)^(sin^2(t)). As t approaches 0, sin^2(t) approaches 0. Therefore, the expression (n)^(sin^2(t)) approaches n^0, which is equal to 1, because any non-zero number raised to the power of 0 is 1.Final Limit Result: Putting it all together, the limit of the given expression as t approaches 0 is 1. This is because the base of the exponentiation approaches n, and the exponent sin^2(t) approaches 0, resulting in n^0, which is 1.

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$\lim_{t \rightarrow 0}(1^{(1/\sin^{2}t)}+2^{(1/\sin^{2}t)}+\dots.+n^{(1/\sin^{2}t)})^{\sin^{2}t}$ is equal to

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\lim_{t \rightarrow 0}(1^{(1/\sin^{2}t)}+2^{(1/\sin^{2}t)}+\dots.+n^{(1/\sin^{2}t)})^{\sin^{2}t}

is equal to

Given Limit Expression: We are given the limit expression $\lim_{t \to 0}(1^{\frac{1}{\sin^2(t)}} + 2^{\frac{1}{\sin^2(t)}} + \ldots + n^{\frac{1}{\sin^2(t)}})^{\sin^2(t)}$ . To solve this, we will use the fact that as $t$ approaches $0$ , $\sin^2(t)$ approaches $0$ as well. We will also use the property that any number raised to the power of $0$ is $1$ .
Consider Base of Exponentiation: First, let's consider the base of the exponentiation, which is $1^{(1/\sin^2(t))} + 2^{(1/\sin^2(t))} + \ldots + n^{(1/\sin^2(t))}$ . As $t$ approaches $0$ , $1/\sin^2(t)$ approaches infinity. Therefore, each term of the form $k^{(1/\sin^2(t))}$ (where $k$ is a constant) approaches $1$ , because any non-zero constant raised to the power of infinity is $1$ .
Sum Approaches $n$ : Since each term in the sum approaches $1$ , the entire sum approaches $n$ , because there are $n$ terms in the sum and each term is approaching $1$ .
Exponent Approaches $0$ : Now, we have the expression $n$ ^( $\sin^2(t)$ ). As $t$ approaches $0$ , $\sin^2(t)$ approaches $0$ . Therefore, the expression $n$ ^( $\sin^2(t)$ ) approaches $n^0$ , which is equal to $1$ , because any non-zero number raised to the power of $0$ is $1$ .
Final Limit Result: Putting it all together, the limit of the given expression as $t$ approaches $0$ is $1$ . This is because the base of the exponentiation approaches $n$ , and the exponent $\sin^2(t)$ approaches $0$ , resulting in $n^0$ , which is $1$ .