Evaluating the limit ${\lim_\limits{x\to0^+}\frac{e^x-\cos(\lambda \sqrt x)}{\sqrt {1+\sin(\lambda x)}-1}}$ 
Evaluate $$\lim_\limits{x\to0^+}\frac{e^x-\cos(\lambda \sqrt x)}{\sqrt {1+\sin(\lambda x)}-1}=(*)$$

My attempt:
I have used Taylor expansion of $e^x, \ \cos x, \ \sin x:$
$$(*)=\lim_\limits{x\to0^+}\frac{1+x-1+\frac{\lambda^2 x}2+ o(x)}{\sqrt {1+\lambda x+o(x)}-1}=\lim_\limits{x\to0^+}\frac{(x+\frac{\lambda^2 x}2)(\sqrt{1+\lambda x+o(x)}+1)}{\lambda x}=\\ =\left(\frac1{\lambda}+\frac{\lambda}2\right)\lim_\limits{x\to0^+}\frac{\sqrt{1+\lambda x}+1}{\lambda x}$$
$\left(\dfrac1{\lambda}+\dfrac{\lambda}2\right)$ is the result written on my textbook, but there seems to be a typo. 
Thanks in advance
P.S. the exercise comes from Calculus Problems, $8.20$ page $144$
EDIT: Actually the last step should have been:
$$\left(\frac1{\lambda}+\frac{\lambda}2\right)\lim_\limits{x\to0^+}\sqrt{1+\lambda x}+1=\frac2{\lambda}+\lambda$$
 A: You are on the right track. At the first step, the $\lambda$ at the numerator    should be squared,
$$\begin{align}
\lim_\limits{x\to0^+}\frac{1+x-1+\frac{\lambda^2 x}2+ o(x)}{\sqrt {1+\lambda x+o(x)}-1}&=\lim_\limits{x\to0^+}\frac{x+\frac{\lambda^2 x}2+ o(x)}{ \lambda x}\cdot \left(\sqrt{1+\lambda x+o(x)}+1\right)\\
&=\left(\frac{1}{\lambda}+\frac{\lambda }2\right)\lim_\limits{x\to0^+}(\sqrt{1+\lambda x+o(x)}+1)=\frac{2}{\lambda}+\lambda
\end{align}$$
which is different from your expected result (probably a typo in your textbook).
A: By standard limits
$$\frac{e^x-\cos(\lambda \sqrt x)}{\sqrt {1+\sin(\lambda x)}-1}=x\frac{\frac{e^x-1}{x}+\frac{1-\cos(\lambda \sqrt x)}{x}}{\sqrt {1+\sin(\lambda x)}-1}\frac{\sqrt {1+\sin(\lambda x)}+1}{\sqrt {1+\sin(\lambda x)}+1}=\left(\sqrt {1+\sin(\lambda x)}+1\right)\left(\frac{e^x-1}{x}+\lambda^2\frac{1-\cos(\lambda \sqrt x)}{\lambda^2x}\right)\frac{\lambda x}{\sin \lambda x}\frac1{\lambda}\\\to2\left(1+\frac12\lambda^2\right)\frac1{\lambda}=\frac2{\lambda}+\lambda$$
A: After expansion and simplification, the dominant terms of the numerator are (the units cancel out)
$$x+\frac{\lambda^2\sqrt x^2}2$$ and that of the denominator (linearize the sine, then the square root, the units also cancel out)
$$\frac{\lambda x}2.$$
Hence the ratio tends to
$$\frac2\lambda+\lambda.$$
