lim _(x arrow 0)(1+x)^(2)/(sin x)=

To find the limit \( \lim_{x \rightarrow 0}(1+x)^{\frac{2}{\sin x}} \), we can use the fact that \( (1+x)^{\frac{1}{x}} \) approaches \( e \) as \( x \) approaches 0.<br /><br />First, let's rewrite the expression in a form that makes it easier to apply this fact:<br /><br />\[ (1+x)^{\frac{2}{\sin x}} = \left( (1+x)^{\frac{1}{\sin x}} \right)^2 \]<br /><br />Now, let \( y = (1+x)^{\frac{1}{\sin x}} \). We want to find the limit of \( y^2 \) as \( x \) approaches 0.<br /><br />Since \( (1+x)^{\frac{1}{x}} \) approaches \( e \) as \( x \) approaches 0, we can write:<br /><br />\[ \lim_{x \rightarrow 0} (1+x)^{\frac{1}{\sin x}} = e \]<br /><br />Therefore,<br /><br />\[ \lim_{x \rightarrow 0} \left( (1+x)^{\frac{1}{\sin x}} \right)^2 = e^2 \]<br /><br />So, the limit is:<br /><br />\[ \lim_{x \rightarrow 0} (1+x)^{\frac{2}{\sin x}} = e^2 \]