Make use of the known limit,
[tex]\displaystyle\lim_{x\to0}\frac{\sin x}x=1[/tex]
We have
[tex]\displaystyle\lim_{x\to0}\frac{e^x\sin x\tan x}{x^2}=\left(\lim_{x\to0}\frac{e^x}{\cos x}\right)\left(\lim_{x\to0}\frac{\sin^2x}{x^2}\right)[/tex]
since [tex]\tan x=\dfrac{\sin x}{\cos x}[/tex], and the limit of a product is the same as the product of limits.
[tex]\dfrac{e^x}{\cos x}[/tex] is continuous at [tex]x=0[/tex], and [tex]\dfrac{e^0}{\cos 0}=1[/tex]. The remaining limit is also 1, since
[tex]\displaystyle\lim_{x\to0}\frac{\sin^2x}{x^2}=\left(\lim_{x\to0}\frac{\sin x}x\right)^2=1^2=1[/tex]
so the overall limit is 1.
