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Evaluate: \(\rm \lim_{x\rightarrow 0}\frac{e^x-(1+x-x^2)}{x^2} = ?\)
1. 1
2. 0
3. 1/2
4. 3/2

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Best answer
Correct Answer - Option 4 : 3/2

Concept:

L-Hospital Rule: Let f(x) and g(x) be two functions

Suppose that we have one of the following cases,

I.  \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{0}{0}\)

II. \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{\infty }{\infty }\)

Then we can apply L-Hospital Rule as:

\(\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = \mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}\)

 Calculation:

\(\rm \lim_{x\rightarrow 0}\frac{e^x-(1+x-x^2)}{x^2} = \frac{0}{0}\)

So, by using L'Hospitals rule i.e. Differentiating numerator and denominator w.r.t. x.

⇒ \(\rm \lim_{x\rightarrow 0}\frac{e^x-(1+x-x^2)}{x^2} = \rm \lim_{x\rightarrow 0}\frac{e^x-(1-2x)}{2x}\)

⇒ \(\rm \lim_{x\rightarrow 0}\frac{e^x-(1-2x)}{2x} = \frac{0}{0}\)

So, again by using L'Hospitals rule i.e. Differentiating numerator and denominator w.r.t. x.

⇒ \(=\rm \lim_{x\rightarrow 0}\frac{e^x+2}{2}\) 

⇒ \(\rm \lim_{x\rightarrow 0}\frac{e^x+2}{2} = \frac{3}{2}\)

Hence, option 4 is the correct answer.

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