Here is limit question\[\lim_{x \rightarrow 0}\left( \frac{ (9^{x}-1)(4^{x}-1) }{ \sqrt{2}-\sqrt{1+\cos(x)}} \right)\]
i hate this type of limit questions
divide numerator and denom. by x^2
lim (a^x-1)/x=ln a
multiply and divide by conjugate : \(\sqrt2+\sqrt{1+cos x}\)
not that way @hartnn
in denominator, 1-cos x/x^2 remains
which is half, and u had ln9.ln4. final answer ln9.ln/root 2 ??
ln9.ln4/root 2 ??
nope
Let me give you the answer first\[8sqrt{2}(ln 3)(ln 2) \]
L'Hopital's ?
nope
ln9.ln4/root 2 is exactly same as 8sqrt2(ln3)(ln2)
i was correct, just needed to simplify
you are not @ hartnn
ln9=2ln 3 ln4 =2ln2 ln9.ln4/root 2 = 4ln 2.ln3 / root 2 = 8 ln2.ln3 *root2 got it ?
nopes, my expression is exactly simplified to your answer.
you are incorrect while you rationalize sqrt 2
can't find how 8 comes......yes, i m getting 2root2 instead of 8root2
then...
let me write it properly again.
\(\huge\frac{4ln3.ln2*2\sqrt2}{1/2}\) i made the mistake of writing 2root 2 in denom. actually its in numerator. now i m getting 16
I'm getting 16 instead of 8 lol
@hartnn yea..
Right, Here is how you solve it @hartnn
me too getting 16
\[\lim_{x \rightarrow 0}\left( \frac{ (9^{x}-1)(4^{x}-1) }{ \sqrt{2}-\sqrt{1+\cos(x)}} \right)\] \[\lim_{x \rightarrow 0}\left( \frac{ (9^{x}-1)(4^{x}-1)times(sqrt{2}+sqrt{1+cos(x)}) }{ 2-1-cos(x)} \right) \] \[\lim_{x \rightarrow 0}\left( \frac{ (9^{x}-1)(4^{x}-1)(sqrt{2}+sqrt{1+cos(x)}) }{ 1-sqrt{1-sin^{2}x }} \right) \] \[\lim_{x \rightarrow 0}( \frac{ (9^{x}-1)(4^{x}-1)(sqrt{2}+sqrt{1+cos(x)})times(1+sqrt{1-sin^{2}x)} }{ sin^{2}x } ) \] \[\lim_{x \rightarrow 0}( (\frac{ x }{ sin(x) })^{2}times(\frac{ 9^{x}-1 }{ x })times(\frac{ 4^{x}-1 }{ x })times(sqrt{2}+sqrt{1+cos(x)})times(1+sqrt{1-sin^{2}x}))\] \[ (1)^{2}times(ln 9)times(ln 4)times(2sqrt{2})times(1) \] \[8sqrt{2}(ln 3)(ln 2) \]
I am closing this
wait how u getting 8, it should be 16
See the steps
u forgot one of the '2's for 1+sqrt(1-sin^2 x)
Ya,ya ...I forget it thanks it 16.
u wrote it as times 1
it should be times 2
It is true
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