$\displaystyle lim_{x->0^{+}}(1+sinx)^{\frac{1}{\sqrt{x}}}$ i get here $\displaystyle 1^{\infty}$ form which states thats its some sort of exponent
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Originally Posted by transgalactic $\displaystyle lim_{x->0^{+}}(1+sinx)^{\frac{1}{\sqrt{x}}}$ i get here $\displaystyle 1^{\infty}$ form which states thats its some sort of exponent $\displaystyle 1^{\infty}$ is one of the indeterminate forms that you can use L'hopital's rule on
lhopital is used on fraction here is exponent
Originally Posted by transgalactic lhopital is used on fraction here is exponent $\displaystyle y = (1+\sin{x})^{\frac{1}{\sqrt{x}}}$ $\displaystyle \ln{y} = \frac{\ln(1+\sin{x})}{\sqrt{x}}}$
why cant i use the exponent definition ?
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