Answer:
[tex]\frac{1}{5}[/tex]
Step-by-step explanation:
Given the expression;
[tex]\lim_{x \to \infty} \dfrac{x+x^2}{5x^2}[/tex]
To find the limit of the function, we need to first divide through by the highest degree if x i.e x² as shown;
[tex]\lim_{x \to \infty} \dfrac{\frac{x}{x^2} +\frac{x^2}{x2}}{\frac{5x^2}{x^2} }\\ \lim_{x \to \infty} \frac{\frac{1}{x}+ 1 }{5}\\As \ x \ tends \ to \ \infty, \ \frac{1}{x} \ tends \ to \ zero. \ Hence;\\ \lim_{x \to \infty} \frac{\frac{1}{x}+ 1 }{5} = \frac{0+1}{5}\\= \dfrac{1}{5}[/tex]
Hence the limit of the function as x tends to infinity is [tex]\frac{1}{5}[/tex]
