# How to solve this limit?#lim_(n->oo)(1+x^n(x^2+4))/(x(x^n+1))#;x>0

To solve the limit (\lim_{n \to \infty} \frac{1+x^n(x^2+4)}{x(x^n+1)}) where (x > 0), we first observe the behavior of the terms as (n) approaches infinity.

Expanding the numerator and denominator, we get: [\lim_{n \to \infty} \frac{1+x^n(x^2+4)}{x(x^n+1)} = \lim_{n \to \infty} \frac{1+x^{n+2}+4x^n}{x^{n+1}+x}]

As (n) tends to infinity, the dominant term in the numerator and the denominator will determine the limit.

Since (x > 0), when (n) goes to infinity, (x^n) will dominate the terms. Therefore, the limit becomes: [\lim_{n \to \infty} \frac{x^n(1+x^2+4)}{x^n(x+1)} = \lim_{n \to \infty} \frac{1+x^2+4}{x+1}]

Now, we can evaluate the limit: [\lim_{n \to \infty} \frac{1+x^2+4}{x+1} = \frac{1+x^2+4}{x+1}]

Thus, the limit of the given expression as (n) approaches infinity is (\frac{1+x^2+4}{x+1}).

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Now, we have to consider 3 Cases :-

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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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