Use the properties of limits to find the limit.
Picture provided below

[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.