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The value of $\underset{x→1}{\lim}\frac{4x^3-x^2+2x-5}{x^6+5x^3-2x-4}$ is: |
$\frac{1}{19}$ $\frac{5}{19}$ $\frac{7}{19}$ $\frac{12}{19}$ |
$\frac{12}{19}$ |
$\underset{x→1}{\lim}\frac{4x^3-x^2+2x-5}{x^6+5x^3-2x-4}$$[\frac{0}{0}form]$ On applying L’Hospital’s rule, we get : $\underset{x→1}{\lim}\frac{12x^2-2x+2}{6x^5+15x^2-2}=\frac{12-2+2}{6+15-2}=\frac{12}{19}$ |
