# $\lim_{h \to 0} \frac{\frac{e^{h \cos \theta(y+h \sin\theta)}-1}{h\cos\theta}-y}{h}$

how do i solve this limit?

$\lim_{h \to 0} \frac{\frac{e^{h \cos \theta(y+h \sin\theta)}-1}{h\cos\theta}-y}{h}$

I tried to do it using equivalents but im not geting to the solution.

Acording to wolfromalfa de correct solution would be

$\displaystyle\frac{y^2}{2}\cos\theta+\sin\theta$

Thanks!

Lets simplify the formula a little bit:

$$=\lim_{h\rightarrow0}\frac{\sec (\theta ) \left(e^{h \cos (\theta ) (h \sin (\theta )+y)}-1\right)-h y}{h^2}$$

then apply two times the l'hospital rule:

$$=\lim_{h\rightarrow0}\frac{\cos (\theta ) \left(4 h^2 \sin ^2(\theta )+4 h y \sin (\theta )+2 \tan (\theta )+y^2\right) e^{h \cos (\theta ) (h \sin (\theta )+y)}}{2}$$

now we can just insert the zero which leads to

$$=\frac{1}{2} \cos (\theta ) \left(2 \tan (\theta )+y^2\right)$$

further simplifing leads to the result from Wolfram-Alpha:

$$=\sin (\theta )+\frac{1}{2} y^2 \cos (\theta )$$

• Thanks!! why is it that if i use equivalents on $e^{h\cos\theta(h\sin\theta+y)}-1$ i cant get to the solution? doesnt it equals to $h\cos\theta(h\sin\theta+y)$? Sep 30, 2016 at 0:42

You can do it with the Taylor expansion of $e^x$.

\begin{align} &\lim_{h\to0}\frac{1}{h}\left(\frac{e^{h\cos\theta(y+h\sin\theta)}-1}{h\cos\theta}-y\right) \\&= \lim_{h\to0}\frac{1}{h}\left(\frac{\sum_{i=0}^\infty\frac{[h\cos\theta(y+h\sin\theta)]^i}{i!}-1}{h\cos\theta}-y\right) \\&= \lim_{h\to0}\frac{1}{h}\left((y+h\sin\theta)+\frac{(h\cos\theta)(y+h\sin\theta)^2}{2}+\ldots-y\right) \\&= \lim_{h\to0}\left(\sin\theta+\frac{\cos\theta(y+h\sin\theta)^2}{2}+\small{\sum_{i=3}^\infty\frac{[h\cos\theta(y+h\sin\theta)]^i}{\cos\theta\cdot i!}}\right) \\&= \sin\theta+\frac{\cos\theta}{2} y^2 \end{align}