`lim_(x->0) tanh(x)/tan(x)` 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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First transform the expression:
`tanh(x)/tan(x) = (sinh(x)/cosh(x)) / (sin(x)/cos(x)) = (sinh(x)/sin(x))*(cos(x)/cosh(x)).`
The second fraction obviously has limit 1 when `x->0`, and we can omit it. The first fraction is equal to
`((sinh(x)-sinh(0))/(x-0)) / ((sin(x)-sin(0))/(x-0)).`
When `x->0,` the numerator tends to sin'(0) and the denominator to sinh'(0), which are cos(0) and cosh(0), respectively, and both are equal to 1.
So the answer is 1.
P.S. Actually we used l’Hospital’s Rule implicitly:)
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