Respuesta :

[tex]\dfrac{4x}{x-5}=\dfrac4{1-\frac5x}[/tex], and as [tex]x\to\infty[/tex], the [tex]\dfrac5x[/tex] term vanishes.

[tex]\dfrac{4x}{x^2+5}=\dfrac{\frac4x}{1+\frac5{x^2}}[/tex], and as [tex]x\to\infty[/tex], both [tex]\dfrac4x[/tex] and [tex]\dfrac5{x^2}[/tex] vanish.

So

[tex]\displaystyle\lim_{x\to\infty}\left(\frac{4x}{x-5}+\frac{4x}{x^2+5}\right)=\lim_{x\to\infty}(4+0)=4[/tex]

and the answer is C.

Answer:

Option C. 4 is the correct option.

Step-by-step explanation:

The given expression is [tex]\lim_{x\rightarrow \oe }[\frac{4x}{x-5}+\frac{4x}{x^{2}+5}][/tex]

We have to calculate the limit of the given expression.

Now we will convert the equation in the form as given below

[tex]=\lim_{x\rightarrow \oe }[\frac{4}{1-\frac{5}{x}}+\frac{4}{x+\frac{5}{x}}][/tex]

We have done this transformation of the equation because we know

[tex]\lim_{x\rightarrow \oe }[\frac{1}{x}]=0[/tex]

Now by the property of limit second term in the question will be vanished

and the remaining part will be

[tex]\lim_{x\rightarrow \oe }[\frac{4}{1-\frac{5}{x}}][/tex]

Now by putting x→∞ we get

[tex][\frac{4}{1-0}]=4[/tex]

Therefore option C is the correct answer.

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