I would like to evaluate the limitlimh→0(1−cos⁡{5h}cos⁡{7h}−1)I'm having a hard time doing this, however. When you try evaluating directly, of course, it comes out in indeterminate form. I tried using the identitylimh→0(1−cos⁡hh)=0by breaking the limit up into two limits, by dividing and multiplying by h, but that only takes me to 0⋅10. What can I do to evaluate this?

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I would like to evaluate the limitlimh→0(1−cos⁡{5h}cos⁡{7h}−1)I&#039;m having a hard time doing this, however. When you try evaluating directly, of course, it comes out in indeterminate form. I tried using the identitylimh→0(1−cos⁡hh)=0by breaking the limit up into two limits, by dividing and multiplying by h, but that only takes me to 0⋅10. What can I do to evaluate this?

muthe2ulj · Accepted Answer

Without L&#039;Hospital:By the double angle formula,1−cos⁡5xcos⁡7x−1=−2sin25x22sin27x2which tends to−(57)2(You can linearize the sines as sin⁡xx tends to 1.)

chabinka61jx · Accepted Answer

Recall the formula 1−cos⁡x2=sin2x2 and the limit limx→0sin⁡xx=1limh→01−cos⁡{5h}cos⁡{7h}−1=limh→02sin25h2−2sin27h2=−limh→0sin25h2sin27h2=−limh→0sin25h2(5h2)2sin25x2(7h2)2⋅(5h2)2(7h2)2=−2549

I would like to evaluate the limit \(\displaystyle\lim_{{{h}\to{0}}}{\left({\frac{{{1}-{\cos{{\left\lbrace{5}{h}\right\rbrace}}}}}{{{\cos{{\left\lbrace{7}{h}\right\rbrace}}}-{1}}}}\right)}\) I'm

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