Correct Answer - Option 3 : 1/4
Concept:
A fraction whose numerator and denominator both tend to zero as x → a is an example of an indeterminate form written as 0/0. it has no definite values. other indeterminate forms are: ∞/∞, ∞ - ∞, 0 x ∞, 1∞, 00, ∞0. indeterminate form are not any definite number and hence are not acceptable as limits. to find limits in such cases, we use the L'hospital's rule, rationalization method, dividing the numerator and denominator by the higher power or factorization method.
Calculation:
\(\rm \lim_{x\rightarrow ∞}[\sqrt{4x^2+x-3}-2x]\), this is ∞ - ∞ form
Here we rationalize numerator
\(\rm \lim_{x\rightarrow ∞}[\sqrt{4x^2+x-3}-2x]\times \frac{\sqrt{4x^2+x-3}+2x}{\sqrt{4x^2+x-3}+2x}\)
\(\rm \lim_{x\rightarrow ∞} \frac{[(\sqrt{4x^2+x-3})^2-(2x)^2]}{\sqrt{4x^2+x-3}+2x}\)
\(\rm \lim_{x\rightarrow ∞} \frac{[{4x^2+x-3}-4x^2]}{\sqrt{4x^2+x-3}+2x}\)
\(\rm \lim_{x\rightarrow ∞} \frac{x-3}{\sqrt{4x^2+x-3}+2x}\), this is ∞/∞ form
Here we divide the numerator and denominator by x
\(\rm \lim_{x\rightarrow ∞} \frac{1-\frac{3}{x}}{\sqrt{4+\frac{1}{x}-\frac{3}{x^2}}+2} = \frac{1}{4}\)
Hence, option 3 is the correct answer.
