IIT JEE , AIEEE Forum - Mathematics , Differential Calculus - Guys, This Is An(supposedly) IIT 2006 Questn.

abrambenny

The Ans. is 1, but i couldn't reach it .I'm gettin plenty of e-s.Pls help

animal

i m not getting ur question can u plz write it in a more clear way plz.

ab_batra

lim sinx^1/x = 0

x-0

bcz it is the infinite power of a number between -1 and 1

lim 1/x^sinx = e ^lim sinx log(1/x)

=e^0=1

hence ans = 0+1=1

thedumbheadwithnobrain

$\mathop{\lim}\limits_{{x} \to {0}}(f(x)+g(x))$ ...... $\mathop{\lim}\limits_{{x} \to {0}}g(x)=\mathop{\lim}\limits_{{x} \to {0}}e^{\frac{log(\frac{1}{x})}{\frac{1}{sinx}}}=\frac{\rightarrow\infty}{\rightarrow\infty}form$ .........applying LH Rule........ $\mathop{\lim}\limits_{{x} \to {0}}e^ { x \frac{sin^2x}{x^2}\frac{1}{cosx}}$ ...... $\mathop{\lim}\limits_{{x} \to {0}}e^{(0)(1)(1)}=\mathop{\lim}\limits_{{x} \to {0}}e^0=1$ ........ $\mathop{\lim}\limits_{{x} \to {0}}f(x)$ ... $\mathop{\lim}\limits_{{x} \to {0}}(m(x))^{n(x)}=\mathop{\lim}\limits_{{x} \to {0}}e^{n(x)(m(x)-1)}$ ... $\mathop{\lim}\limits_{{x} \to {0}}e^{\frac{sinx-1}{x}}$ ...... $\mathop{\lim}\limits_{{x} \to {0}}e^{\rightarrow (-{\infty})}=0$ .......... $Hence\;answer\;is\;1$ ,,,,cheers

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