Answer:
Step-by-step explanation:
Given expression is,
[tex]\lim_{m \to \infty} \frac{4m^2-6m+1}{3m^2+5m+7}[/tex]
By dividing denominator and numerator of the given fraction by m²,
[tex]\lim_{m \to \infty} \frac{\frac{4m^2-6m+1}{m^2} }{\frac{3m^2+5m+7}{m^2} }[/tex]
= [tex]\lim_{m \to \infty} \frac{4-\frac{6}{m} +\frac{1}{m^2} }{3+\frac{5}{m} +\frac{7}{m^2} }[/tex]
We know that [tex]\lim_{n \to \infty} \frac{1}{n}=0[/tex]
[tex]\lim_{m \to \infty} \frac{4-\frac{6}{m} +\frac{1}{m^2} }{3+\frac{5}{m} +\frac{7}{m^2} }=\frac{4-0+0}{3+0+0}[/tex]
[tex]=\frac{4}{3}[/tex]
Therefore, We can evaluate the limit of the expression by substituting ∞ in for m.
