MCQ
Let $f(x) =$ $\frac{{\ln ({x^2} + {e^x})}}{{\ln ({x^4} + {e^{2x}})}}$ . If $f(x) = l$ and $\mathop {Limit}\limits_{x\,\, \to \,\, - \,\infty } f(x) = m$ then :
- ✓$l = m$
- B$l = 2m$
- C$2 l = m$
- D$l + m = 0$
= $\mathop {Lim}\limits_{x \to \infty } \,$ $\frac{{\ln \,\,{e^x}\,\left( {1 + \frac{{{x^2}}}{{{e^x}}}} \right)}}{{\ln \,\,{e^{2x}}\,\left( {1 + \frac{{{x^4}}}{{{e^{2x}}}}} \right)}}$
=$\mathop {Lim}\limits_{x \to \infty } \,$ $\frac{{x\, + \,\ln \,\,\left( {1 + \frac{{{x^2}}}{{{e^x}}}} \right)}}{{2x\, + \,\ln \,\,\,\left( {1 + \frac{{{x^4}}}{{{e^{2x}}}}} \right)}}$
= $\frac{{1\, + \,\ln \,\,{{\left( {1 + \frac{{{x^2}}}{{{e^x}}}} \right)}^{1/x}}}}{{2\, + \,\ln \,\,\,{{\left( {1 + \frac{{{x^4}}}{{{e^{2x}}}}} \right)}^{1/x}}}}$
note that $\mathop {as}\limits_{x \to \,\infty } \,$ $\,\frac{{{x^2}}}{{{e^x}}}\,\, \to \,\,0$ and $\mathop {as}\limits_{x \to \,\infty } \,$ $\,\frac{{{x^2}}}{{{e^{2x}}}}\,\, \to \,\,0$
=$\mathop {Lim}\limits_{x \to \infty } \,$ $\frac{{1 + \frac{1}{x}\left( {1 + \frac{{{x^2}}}{{{e^x}}}\, - \,1} \right)}}{{2 + \frac{1}{x}\left( {1 + \frac{{{x^4}}}{{{e^{2x}}}}\, - \,1} \right)}}$
= $\mathop {Lim}\limits_{x \to \infty } \,$ $\frac{{1 + \frac{x}{{{e^x}}}}}{{2 + \frac{{{x^3}}}{{{e^{2x}}}}}}$
|||$^{ly}$ $\mathop {Limit}\limits_{x \to \, - \,\infty } \,\, = \,\,\frac{1}{2}\,$
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