Question

plz solve part 3 of this question...​

Answer 1

Answer:

$\huge\bold\red{\frac{3}{2}}$

Step-by-step explanation:

Given,

$(iii) \lim_{x \to 0 } \frac{ {e}^{ {x}^{2} } - \cos(x) }{ {x}^{2} }$

Now,

Applying L- Hospital Rule,

we get,

$= \lim_{x \to 0 } \frac{( {e}^{ {x}^{2} } \times 2x )+ \sin(x) }{2x}$

Now,

simplifying on the terms,

we get,

$= \lim_{x \to 0 } {e}^{ {x}^{2} } + \lim_{x \to 0 } \frac{ \sin(x) }{2x}$

But,

we know that,

$\lim_{x \to 0 } \frac{ \sin(x) }{x} = 1$

Therfore,

pUtting the values,

we get,

$= {e}^{0} + \frac{1}{2} \\ \\ = 1 + \frac{1}{2} \\ \\ = \frac{3}{2}$

Hence,

$\large\bold{Value = \frac{3}{2}}$