The mathematical expression in the image reads:

$\lim_{x \to 0} \left( 1 - e^{\tan^2\left(\sqrt{x}\right)} \right)^{\frac{1}{\sqrt[3]{x^2}}}.$

Let’s solve it step by step:

Step 1: Analyze the Limit

We are taking a limit as $x \to 0$. Inside the expression:

1. $\tan^2(\sqrt{x}) \to 0$ as $x \to 0$, since $\tan(0) = 0$.
2. This implies $e^{\tan^2(\sqrt{x})} \to e^0 = 1$.
3. Therefore, $1 - e^{\tan^2(\sqrt{x})} \to 0$.

However, the exponent $\frac{1}{\sqrt[3]{x^2}}$ diverges to $\infty$ as $x \to 0^+$. This results in an indeterminate form of the type $0^\infty$. To resolve this, we apply logarithmic techniques.

Step 2: Apply the Natural Logarithm

Let $L = \lim_{x \to 0} \left( 1 - e^{\tan^2(\sqrt{x})} \right)^{\frac{1}{\sqrt[3]{x^2}}}$.

Take the natural logarithm of both sides: $\ln L = \lim_{x \to 0} \frac{\ln\left( 1 - e^{\tan^2(\sqrt{x})} \right)}{\sqrt[3]{x^2}}.$

Now, the problem reduces to evaluating the limit of the logarithmic term: $\ln\left(1 - e^{\tan^2(\sqrt{x})}\right).$

Step 3: Expand Using Approximation

For small $x$, we use the series expansions:

1. $\tan^2(\sqrt{x}) \approx \sqrt{x}^2 = x$, since $\tan(z) \approx z$ for small $z$.
2. $e^{\tan^2(\sqrt{x})} \approx e^x \approx 1 + x$ for small $x$.
3. Therefore, $1 - e^{\tan^2(\sqrt{x})} \approx 1 - (1 + x) = -x$.

Substitute into the logarithm: $\ln\left(1 - e^{\tan^2(\sqrt{x})}\right) \approx \ln(-x).$

Step 4: Substitute and Simplify

Rewrite the limit: $\ln L = \lim_{x \to 0} \frac{\ln(-x)}{\sqrt[3]{x^2}}.$

(At this step, we would need to ensure if $x \to 0^+$ or $x \to 0^-$, as negative logarithms arise. Would you like to clarify?)

Let me know how much detail you'd like to proceed further or if clarification is needed!

5 Questions for Deeper Understanding:

1. What is the significance of resolving $0^\infty$ indeterminate forms in limits?
2. How does the logarithmic method simplify exponentiation limits?
3. Why does $\tan^2(\sqrt{x})$ simplify to $x$ for small values of $x$?
4. What happens if $x$ approaches $0$ from the negative side instead?
5. Can we use L’Hôpital’s Rule in this problem? If yes, how?

Tip: When dealing with exponential limits, always consider logarithmic simplifications to reduce complexity.