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

Answer 1

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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Answer 2

#"The Reqd. Lim.="1/x, if 0 lt x lt 1;#
#=3, if x=1; and,#
#=x+4/x, if 1 lt x.#

#"Reqd. Lim."=lim_(n to oo) {1+x^n(x^2+4)}/{x(x^n+1)}#
#=lim_(n to oo) {1+x^(n+2)+4x^n}/{x^(n+1)+x}......(1)#
#=lim_(n to oo) {x^n(1/x^n+x^2+4)}/{x^n(x+1/x^(n-1))#
#=lim_(n to oo) (1/x^n+x^2+4)/(x+1/x^(n-1))......(2).#

Now, we have to consider 3 Cases :-

Case 1 : #0 lt x lt 1.#
In this Case , we know that, as #n to oo, x^n to 0.#
And, #"by (1), :., Reqd. Lim.="{1+x^2(0)+4(o)}/{x(0)+x)=1/x.#
Case 2 : #x=1.#
#"The Reqd. Lim.="(1+1+4)/(1+1)=6/2=3.#
Case 3 : #x gt 1.#
In this Case, # because 0 lt 1/x lt 1, :. " as "n to oo, 1/x^n to 0.#
Hence, #"by (2), the Reqd. Lim."=(0+x^2+4)/{x+1/x(0)}=(x^2+4)/x, or, x+4/x.#

Enjoy Maths.!

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Answer 3

See below.

Considering #x ne 0#
If #abs x > 1#
#lim_(n->oo)(1+x^n(x^2+4))/(x(x^n+1))=lim_(n->oo)x^n/x^n((1/x^n+(x^2+4)))/(x(1+1/x^n)) = (x^2+4)/x#
If #abs x < 1#
#lim_(n->oo)(1+x^n(x^2+4))/(x(x^n+1))=1/x#
If #x = 1#
#lim_(n->oo)(1+x^n(x^2+4))/(x(x^n+1))=3#
If #x = -1#
#lim_(n->oo)(1+x^n(x^2+4))/(x(x^n+1))# does not exists
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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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