Correct Answer - Option 1 : 2
Concept:
L-Hospital Rule: Let f(x) and g(x) be two functions
Suppose that we have one of the following cases,
i. \(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \;\frac{0}{0}\)
ii. \(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \;\frac{\infty }{\infty }\)
Then we can apply L-Hospital Rule ⇔ \(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \;\mathop {\lim }\limits_{x \to a} \frac{{f'\left( x \right)}}{{g'\left( x \right)}}\)
Note: We have to differentiate both the numerator and denominator with respect to x unless and until \(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = l \ne \frac{0}{0}\)where l is a finite value.
Calculation:
\(\rm \mathop {\lim }\limits_{x\rightarrow \infty }\frac{2x^{2}+3}{x^{2}+1}\)
Apply L-Hospital rule, we get
\(=\rm \mathop {\lim }\limits_{x\rightarrow \infty }\frac{\frac{\mathrm{d} (2x^{2}+3)}{\mathrm{d} x}}{\frac{\mathrm{d} (x^{2}+1)}{\mathrm{d} x}}\)
\(=\rm \mathop {\lim }\limits_{x\rightarrow \infty }\frac{4x}{2x}\)
\(\therefore\rm \mathop {\lim }\limits_{x\rightarrow \infty }\frac{2x^{2}+3}{x^{2}+1}=2\)
Hence , option 1 is correct.
