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Question: \(\lim_{x \rightarrow 0}\frac{1}{x^{5}}\int_{0}^{x}e^{- t^{2}}\) dt – \(\frac{1}{x^{4}} + \frac{1}{3...

lim⁡x→01x5∫0xe−t2\lim_{x \rightarrow 0}\frac{1}{x^{5}}\int_{0}^{x}e^{- t^{2}} dt – 1x4+13x2\frac{1}{x^{4}} + \frac{1}{3x^{2}} is equal to

A

1

B

3/5

C

½

D

3/10

Answer

3/10

Explanation

Solution

Applying L'Hospital Rule

3∫0xe−t2dt−3x+x3x5\frac{3\int_{0}^{x}e^{- t^{2}}dt - 3x + x^{3}}{x^{5}} (00)\left( \frac{0}{0} \right)

lim⁡x→0\lim _ { x \rightarrow 0 } 3e−x2−3+3x25x4\frac{3e^{- x^{2}} - 3 + 3x^{2}}{5x^{4}}

= 35\frac{3}{5} lim⁡x→0\lim_{x \rightarrow 0} e−x2−1+x2x4\frac{e^{- x^{2}} - 1 + x^{2}}{x^{4}} (00)\left( \frac{0}{0} \right)

= 35\frac{3}{5} lim⁡x→0\lim_{x \rightarrow 0} e−x2(−2x)−0+2x4x3\frac{e^{- x^{2}}( - 2x) - 0 + 2x}{4x^{3}}

= 35⋅24\frac { 3 } { 5 } \cdot \frac { 2 } { 4 } lim⁡x→0\lim_{x \rightarrow 0} −e−x2+1x2\frac{- e^{- x^{2}} + 1}{x^{2}} (00)\left( \frac{0}{0} \right)

= 35×24\frac { 3 } { 5 } \times \frac { 2 } { 4 } lim⁡x→0\lim_{x \rightarrow 0} 1−e−x2(−x2)\frac{1 - e^{- x^{2}}}{( - x^{2})} = 310\frac{3}{10}