Calculating the Limit of $\displaystyle \frac{5^{\sin{h}}-1}{\tan{h}}$

[Guest2]

With this one I don't know where to start

$\displaystyle \lim_{h \to 0} \frac{5^{\sin{h}}-1}{\tan{h}}$

 

[MarkFL]

With this one I don't know where to start

$\displaystyle \lim_{h \to 0} \frac{5^{\sin{h}}-1}{\tan{h}}$

I would use L'Hôpital's Rule...:)

 

[Guest2]

I would use L'Hôpital's Rule...:)

Is this correct?

$ \ell :=\displaystyle \lim_{h \to 0} \frac{5^{\sin{h}}-1}{\tan{h}} $ let $t = \tan{h}$ then $ \sin(h) = \sin(\arctan{t}) = \frac{t}{\sqrt{t^2+1}}$.

$$\begin{aligned} \displaystyle \ell & = \lim_{t \to 0} \frac{5^{\dfrac{t}{\sqrt{t^2+1}}}-1}{t} \\& =\lim_{t \to 0} \frac{5^{\dfrac{t+x}{\sqrt{(t+x)^2+1}}}-5^{\dfrac{x}{\sqrt{x^2+1}}}}{t}\bigg|_{x=0} \\& = \frac{d}{dx} 5^{\dfrac{x}{\sqrt{x^2+1}}}\bigg|_{x=0} \\& = \frac{5^{\dfrac{x}{\sqrt{x^2+1}}}\log(5)}{\sqrt{(x^2+1)^3}}\bigg|_{x=0} \\& = \log(5). \end{aligned}$$

 

[Guest2]

Never mind, L'hopital is indeed the best way to do this.