Notes: `lim_(x->oo)x^3/e^(x^2)` Evaluate the limit, using L’Hôpital’s Rule if necessary.

Tuesday, 28 October 2014

$\displaystyle\lim_{{{x}\to\infty}}\frac{{{x}}^{{3}}}{{{e}}^{{{{x}}^{{2}}}}}$ Evaluate the limit, using L’Hôpital’s Rule if necessary.

Given to solve ,

$\displaystyle\lim_{{{x}\to\infty}}\frac{{{x}}^{{3}}}{{{e}}^{{{{x}}^{{2}}}}}$

upon $\displaystyle{x}$ tends to $\displaystyle\infty$ we get $\displaystyle\frac{{{x}}^{{3}}}{{{e}}^{{{{x}}^{{2}}}}}=\frac{\infty}{\infty}$

so, by applying the L'Hopital Rule we get

as for the general equation it is as follows

$\displaystyle\lim_{{{x}\to{a}}}\frac{{f{{\left({x}\right)}}}}{{g{{\left({x}\right)}}}}{i}{s}=\frac{{0}}{{0}}$ or $\displaystyle\frac{{\pm\infty}}{{\pm\infty}}$ then by using the L'Hopital Rule we get the solution with the below form.

$\displaystyle\lim_{{{x}\to{a}}}\frac{{{f{'}}{\left({x}\right)}}}{{{g{'}}{\left({x}\right)}}}$

so , now evaluating

$\displaystyle\lim_{{{x}\to\infty}}\frac{{{x}}^{{3}}}{{{e}}^{{{{x}}^{{2}}}}}$

upon using the L'Hopital Rule

= $\displaystyle\lim_{{{x}\to\infty}}\frac{{{\left({{x}}^{{3}}\right)}'}}{{{\left({{e}}^{{{{x}}^{{2}}}}\right)}'}}$

= $\displaystyle\lim_{{{x}\to\infty}}\frac{{{3}{{x}}^{{2}}}}{{{{e}}^{{{{x}}^{{2}}}}{\left({2}{x}\right)}}}$

=> $\displaystyle\lim_{{{x}\to\infty}}\frac{{{3}{x}}}{{{{e}}^{{{{x}}^{{2}}}}{\left({2}\right)}}}$

now...

Given to solve ,

$\displaystyle\lim_{{{x}\to\infty}}\frac{{{x}}^{{3}}}{{{e}}^{{{{x}}^{{2}}}}}$

upon $\displaystyle{x}$ tends to $\displaystyle\infty$ we get $\displaystyle\frac{{{x}}^{{3}}}{{{e}}^{{{{x}}^{{2}}}}}=\frac{\infty}{\infty}$

so, by applying the L'Hopital Rule we get

as for the general equation it is as follows

$\displaystyle\lim_{{{x}\to{a}}}\frac{{{f{'}}{\left({x}\right)}}}{{{g{'}}{\left({x}\right)}}}$

so , now evaluating

$\displaystyle\lim_{{{x}\to\infty}}\frac{{{x}}^{{3}}}{{{e}}^{{{{x}}^{{2}}}}}$

upon using the L'Hopital Rule

= $\displaystyle\lim_{{{x}\to\infty}}\frac{{{\left({{x}}^{{3}}\right)}'}}{{{\left({{e}}^{{{{x}}^{{2}}}}\right)}'}}$

= $\displaystyle\lim_{{{x}\to\infty}}\frac{{{3}{{x}}^{{2}}}}{{{{e}}^{{{{x}}^{{2}}}}{\left({2}{x}\right)}}}$

=> $\displaystyle\lim_{{{x}\to\infty}}\frac{{{3}{x}}}{{{{e}}^{{{{x}}^{{2}}}}{\left({2}\right)}}}$

now on $\displaystyle{x}\to\infty$ we get $\displaystyle\frac{{{3}{x}}}{{{{e}}^{{{{x}}^{{2}}}}{\left({2}\right)}}}=\frac{\infty}{\infty}$

so,again by applying the L'Hopital Rule we get

$\displaystyle\lim_{{{x}\to\infty}}\frac{{{3}{x}}}{{{{e}}^{{{{x}}^{{2}}}}{\left({2}\right)}}}$

= $\displaystyle\lim_{{{x}\to\infty}}\frac{{{\left({3}{x}\right)}'}}{{{\left({{e}}^{{{{x}}^{{2}}}}{\left({2}\right)}\right)}'}}$

= $\displaystyle\lim_{{{x}\to\infty}}\frac{{{3}}}{{{{e}}^{{{{x}}^{{2}}}}{\left({2}\right)}{\left({2}{x}\right)}}}$

= $\displaystyle\lim_{{{x}\to\infty}}\frac{{{3}}}{{{{e}}^{{{{x}}^{{2}}}}{\left({4}{x}\right)}}}$

now as $\displaystyle{x}\to\infty$

$\displaystyle\frac{{3}}{{{{e}}^{{{{x}}^{{2}}}}{\left({4}{x}\right)}}}=\frac{{3}}{{{{e}}^{{{\infty}^{{2}}}}{\left({4}{\left(\infty\right)}\right)}}}={0}$

$\displaystyle\lim_{{{x}\to\infty}}\frac{{{3}}}{{{\left({{e}}^{{{{x}}^{{2}}}}{\left({4}{x}\right)}\right)}}}$ $\displaystyle={0}$

now we can state that

$\displaystyle\lim_{{{x}\to\infty}}\frac{{{x}}^{{3}}}{{{e}}^{{{{x}}^{{2}}}}}$ $\displaystyle={0}$

Notes

Tuesday, 28 October 2014

$\displaystyle\lim_{{{x}\to\infty}}\frac{{{x}}^{{3}}}{{{e}}^{{{{x}}^{{2}}}}}$ Evaluate the limit, using L’Hôpital’s Rule if necessary.

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How are race, gender, and class addressed in Oliver Optic's Rich and Humble?

Tuesday, 28 October 2014

Evaluate the limit, using L’Hôpital’s Rule if necessary.

No comments:

Post a Comment

How are race, gender, and class addressed in Oliver Optic&#39;s Rich and Humble?

$\displaystyle\lim_{{{x}\to\infty}}\frac{{{x}}^{{3}}}{{{e}}^{{{{x}}^{{2}}}}}$ Evaluate the limit, using L’Hôpital’s Rule if necessary.

How are race, gender, and class addressed in Oliver Optic's Rich and Humble?