Evaluate the limit as x approaches 0 from the right of (x / ((e^x - 1) cos root x)) raised to the power 1 / ((1 + sin(x^2))^(1/x) - 1). Choose from e^(1/2), e^(-1/2), e or 1.

Step-by-step solution

1. 1

   Identify the indeterminate form and set up the standard reduction. The base tends to $1$ and the exponent grows without bound, so this is $1^{\infty}$. Write $L=\lim u^{v}$ and use

   $L=\exp\left(\lim v\ln u\right)\approx\exp\left(\lim v(u-1)\right),$

   since $\ln u\sim u-1$ when $u\to 1$. Everything now depends on how fast $u-1\to 0$ compared with $v\to\infty$.

2. 2

   Expand the exponent's denominator. For small $x$, $\sin x^{2}=x^{2}+O(x^{6})$, so

   $\left(1+\sin x^{2}\right)^{1/x}=\exp\left(\frac{\ln\left(1+x^{2}+\cdots\right)}{x}\right)=\exp\left(x+O(x^{3})\right)=1+x+O(x^{2}).$

   Subtracting $1$ leaves $x+O(x^2)$, so the exponent is

   $v=\frac{1}{x+O(x^{2})}\sim\frac{1}{x}.$

3. 3

   Expand the base to enough orders. Use $e^{x}-1=x+\tfrac{x^{2}}{2}+O(x^{3})$ and $\cos\sqrt{x}=1-\tfrac{x}{2}+O(x^{2})$. Their product is

   $\left(x+\tfrac{x^{2}}{2}\right)\left(1-\tfrac{x}{2}\right)=x+\tfrac{x^{2}}{2}-\tfrac{x^{2}}{2}+O(x^{3})=x+O(x^{3}).$

   The two first-order corrections cancel exactly — that cancellation is the whole point of the problem.

4. 4

   Form the base and see how close to 1 it is.

   $u=\frac{x}{x+O(x^{3})}=1+O(x^{2}),\qquad\text{so}\qquad u-1=O(x^{2}).$

5. 5

   Combine exponent and base.

   $v(u-1)=\frac{1}{x}\cdot O\!\left(x^{2}\right)=O(x)\longrightarrow 0,$

   hence

   $L=e^{0}=1.$

   Had the $x^2$ terms not cancelled in step 3, $u-1$ would have been of order $x$ and the answer would have been $e^{\pm 1/2}$ — which is exactly what options (A) and (B) are baiting.

6. 6

   Confirm numerically. Evaluating the original expression gives $1.004211$ at $x=0.1$, $1.000417$ at $x=0.01$, $1.0000417$ at $x=0.001$ and $1.0000042$ at $x=10^{-4}$ — approaching $1$ linearly in $x$, exactly as the $O(x)$ estimate predicts. The answer is (D).