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L'Hospital's rule with trigs

Hello all!

I need some help with this question:

Attachment 25693

the rule requires f(x)/g(x) to go to an indeterminate form, does that mean I rewrite this as a quotient?

so lim as x-> 0 of the xth root of $\displaystyle (sinx+cosx)$

is that right?

Re: L'Hospital's rule with trigs

Quote:

Originally Posted by

**kingsolomonsgrave** Hello all!

I need some help with this question:

Attachment 25693
the rule requires f(x)/g(x) to go to an indeterminate form, does that mean I rewrite this as a quotient?

so lim as x-> 0 of the xth root of $\displaystyle (sinx+cosx)$

is that right?

if

$\displaystyle y=\lim_{x->0}[cos(x)+sin(x)]^{1/x}$

so: $\displaystyle ln(y)=\lim_{x->0}ln[[cos(x)+sin(x)]^{1/x}]$

Now you most use the logarithmic power rule and you will get indeterminate form: 0/0

Re: L'Hospital's rule with trigs

$\displaystyle ln(y)=$lim as x->0$\displaystyle ln(sinx/x + cosx/x)$

am I on the right track?

Re: L'Hospital's rule with trigs

Quote:

Originally Posted by

**kingsolomonsgrave** $\displaystyle ln(y)=$lim as x->0$\displaystyle ln(sinx/x + cosx/x)$

am I on the right track?

This is not true you cant bring the $\displaystyle \frac{1}{x} $ "into" your ln function. You want to write $\displaystyle \frac{\ln( \sin(x) + \cos(x))}{x} $