Indirect limits

As $x \to 0 \Rightarrow e^x - 1 \to x \rightarrow$ by expansion.

Similarly,

As $x \to 0 \Rightarrow sin(\frac{x^2}{\lambda^2}) \to \frac{x^2}{\lambda^2} \rightarrow$ by expansion

and

As $x \to 0 \Rightarrow ln(1+\frac{x^2}{2}) \to \frac{x^2}{2} \rightarrow$ by expansion.

For $f(x)$ to be continuous:

$\displaystyle \lim_{x \to 0} f(x) = f(0)$

$\displaystyle \lim_{x \to 0} \dfrac{x^4}{\frac{x^2} {\lambda^2} \frac{x^2}{2}} = 2\lambda^2$

$2\lambda^2 = 8 \Rightarrow \lambda^2 = 4$

People expect all these $\lambda$ to be a positive quantity. So at first look after finding the answer, the candidate must get confused. In order to confuse the candidate, the square of the answer will also be given, so that the candidate marks the wrong option thinking the answer has to be positive.

$(-2)^2=4$ , the answer is $-2$ .

Not limits, human psychology XD

Since the given function is indeterminate at 0, we could use L'Hôpital's rule. The following rule for limits is especially important if you want to reduce the complexity of your work:

$\lim _{ x\rightarrow a }{ \, f(x)\, \cdot \, g(x)\, =\, } \lim _{ x\rightarrow a }{ \, f(x)\, \cdot \, \lim _{ x\rightarrow a }{ \, g(x) } }$

Then we can split it into two parts:

$\lim _{ x\rightarrow 0 }{ \frac { { \left( { e }^{ x }\, -\, 1 \right) }^{ 4 } }{ \sin { \left( \frac { { x }^{ 2 } }{ { \lambda }^{ 2 } } \right) } \ln { \left( 1\, +\, \frac { { x }^{ 2 } }{ 2 } \right) } } } \, =\, \lim _{ x\rightarrow 0 }{ \frac { { \left( { e }^{ x }\, -\, 1 \right) }^{ 2 } }{ \sin { \left( \frac { { x }^{ 2 } }{ { \lambda }^{ 2 } } \right) } } } \cdot \lim _{ x\rightarrow 0 }{ \frac { { \left( { e }^{ x }\, -\, 1 \right) }^{ 2 } }{ \ln { \left( 1\, +\, \frac { { x }^{ 2 } }{ 2 } \right) } } } \, \\ \Longrightarrow \, \lim _{ x\rightarrow 0 }{ { \frac { { \lambda }^{ 2 }\left( { e }^{ x }\, -\, 1 \right) }{ x } } } \cdot \lim _{ x\rightarrow 0 }{ \frac { 2\left( { e }^{ x }\, -\, 1 \right) }{ x } } \\ \Longrightarrow \, \lim _{ x\rightarrow 0 }{ \left( { \lambda }^{ 2 } \right) } \cdot \lim _{ x\rightarrow 0 }{ 2 } \, =\, 2{ \lambda }^{ 2 }\, =\, 8\\ \Longrightarrow \, { \lambda }\, =\, \pm 2$