Could somebody help me to solve this, please?

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Answer 1 · 2017-10-11T10:20:24

See below.

#lim_(h->0)(cos(h)-1)/h#

We need to use L'Hopital's rule for this since we have the indeterminate form #0/0#.

Differentiate numerator and denominator.

#d/(dh) cos(h)-1=d/(dh)cos(h)-d/(dh)1#

#-> d/(dh)cos(h)-d/(dh)1=-sin(h)-0#

#->-sin(h)#

And denominator:

#d/(dh)x=1#

So:

#lim_(h->0)(-sin(h))/1#

Plugging in #h=0#

#lim_(h->0)(-sin(0))/1= 0/1=0#

Therefore:

#lim_(h->0)(cos(h)-1)/h=0#