`lim_(x->(pi/2)^-) cos(x) sec(5x)` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t apply, explain why.
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Here we can use more elementary method.
I suppose that we know `lim_(x->0)(sinx/x) =1.`
Let's substitute `x=pi/2-y,` then `y->0` .
`cosx*sec(5x) = cos(x)/cos(5x) = cos(pi/2-y)/cos((5pi)/2-5y) = siny/sin(5y).`
This, in turn, = `(1/5)*(siny/y)/(sin(5y)/(5y)),`
which has limit 1/5 when `y->0` (if we want we may also sustitute 5y=z->0 for the denominator).
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