# `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,...

`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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### 1 Answer

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:)