limit of exponent type

**transgalactic** · September 11th 2011, 10:25 AM

$lim_{x->0^{+}}(1+sinx)^{\frac{1}{\sqrt{x}}}$

i get here $1^{\infty}$ form which states thats its some sort of exponent

**e^(i*pi)** · September 11th 2011, 10:31 AM

Originally Posted by transgalactic

$lim_{x->0^{+}}(1+sinx)^{\frac{1}{\sqrt{x}}}$

i get here $1^{\infty}$ form which states thats its some sort of exponent

$1^{\infty}$ is one of the indeterminate forms that you can use L'hopital's rule on

**transgalactic** · September 11th 2011, 10:33 AM

lhopital is used on fraction here is exponent

**skeeter** · September 11th 2011, 10:45 AM

Originally Posted by transgalactic

lhopital is used on fraction here is exponent

$y = (1+\sin{x})^{\frac{1}{\sqrt{x}}}$

$\ln{y} = \frac{\ln(1+\sin{x})}{\sqrt{x}}}$

**transgalactic** · September 11th 2011, 10:47 AM

why cant i use the exponent definition
?

Thread: limit of exponent type