Notes: `lim_(x->oo)x^2/sqrt(x^2+1)` Evaluate the limit, using L’Hôpital’s Rule if necessary.

Tuesday, 5 September 2017

$\displaystyle\lim_{{{x}\to\infty}}\frac{{{x}}^{{2}}}{\sqrt{{{{x}}^{{2}}+{1}}}}$ Evaluate the limit, using L’Hôpital’s Rule if necessary.

Given

$\displaystyle\lim_{{{x}\to\infty}}\frac{{{x}}^{{2}}}{\sqrt{{{{x}}^{{2}}+{1}}}}$

as $\displaystyle{x}\to\infty$ then we get $\displaystyle\frac{{{x}}^{{2}}}{\sqrt{{{{x}}^{{2}}+{1}}}}=\frac{\infty}{\infty}$

since it is of the form $\displaystyle\frac{\infty}{\infty}$ , we can use the L 'Hopital rule

so upon applying the L 'Hopital rule we get the solution as follows,

For the given general equation L 'Hopital rule 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...

Given

$\displaystyle\lim_{{{x}\to\infty}}\frac{{{x}}^{{2}}}{\sqrt{{{{x}}^{{2}}+{1}}}}$

as $\displaystyle{x}\to\infty$ then we get $\displaystyle\frac{{{x}}^{{2}}}{\sqrt{{{{x}}^{{2}}+{1}}}}=\frac{\infty}{\infty}$

since it is of the form $\displaystyle\frac{\infty}{\infty}$ , we can use the L 'Hopital rule

so upon applying the L 'Hopital rule we get the solution as follows,

For the given general equation L 'Hopital rule is as follows

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

so , now evaluating

$\displaystyle\lim_{{{x}\to\infty}}\frac{{{x}}^{{2}}}{\sqrt{{{{x}}^{{2}}+{1}}}}$

= $\displaystyle\lim_{{{x}\to\infty}}\frac{{{\left({{x}}^{{2}}\right)}'}}{{{\left(\sqrt{{{{x}}^{{2}}+{1}}}\right)}'}}$

First let us solve $\displaystyle{\left(\sqrt{{{{x}}^{{2}}+{1}}}\right)}'$

=> $\displaystyle\frac{{d}}{{\left.{d}{x}}}{\left(\sqrt{{{{x}}^{{2}}+{1}}}\right)}$

let $\displaystyle{u}={{x}}^{{2}}+{1}$

so,

$\displaystyle\frac{{d}}{{\left.{d}{x}}}{\left(\sqrt{{{{x}}^{{2}}+{1}}}\right)}$

= $\displaystyle\frac{{d}}{{\left.{d}{x}}}{\left(\sqrt{{{u}}}\right)}$

= $\displaystyle\frac{{d}}{{{d}{u}}}\sqrt{{{u}}}\cdot\frac{{d}}{{\left.{d}{x}}}{\left({u}\right)}$ [as $\displaystyle\frac{{d}}{{\left.{d}{x}}}{f{{\left({u}\right)}}}=\frac{{d}}{{{d}{u}}}{f{{\left({u}\right)}}}\frac{{{d}{u}}}{{\left.{d}{x}}}$ ]

= $\displaystyle{\left[{\left(\frac{{1}}{{2}}\right)}{{u}}^{{{\left(\frac{{1}}{{2}}\right)}-{1}}}\right]}\cdot{\left(\frac{{d}}{{\left.{d}{x}}}{\left({{x}}^{{2}}+{1}\right)}\right)}$

= $\displaystyle{\left[{\left(\frac{{1}}{{2}}\right)}{{u}}^{{-\frac{{1}}{{2}}}}\right]}\cdot{\left({2}{x}\right)}$

= $\displaystyle{\left[\frac{{1}}{{{2}\sqrt{{{{x}}^{{2}}+{1}}}}}\right]}\cdot{\left({2}{x}\right)}$

= $\displaystyle\frac{{x}}{\sqrt{{{{x}}^{{2}}+{1}}}}$

so now the below limit can be given as

= $\displaystyle\lim_{{{x}\to\infty}}\frac{{{\left({{x}}^{{2}}\right)}'}}{{{\left(\sqrt{{{{x}}^{{2}}+{1}}}\right)}'}}$

= $\displaystyle\lim_{{{x}\to\infty}}\frac{{{\left({2}{x}\right)}}}{{\frac{{x}}{\sqrt{{{{x}}^{{2}}+{1}}}}}}$

= $\displaystyle\lim_{{{x}\to\infty}}{\left({2}\sqrt{{{{x}}^{{2}}+{1}}}\right)}$

Now on substituting the value of $\displaystyle{x}=\infty$ we get

= $\displaystyle{\left({2}\sqrt{{{{\left(\infty\right)}}^{{2}}+{1}}}\right)}$

= $\displaystyle\infty$

Notes

Tuesday, 5 September 2017

$\displaystyle\lim_{{{x}\to\infty}}\frac{{{x}}^{{2}}}{\sqrt{{{{x}}^{{2}}+{1}}}}$ 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, 5 September 2017

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?

$\displaystyle\lim_{{{x}\to\infty}}\frac{{{x}}^{{2}}}{\sqrt{{{{x}}^{{2}}+{1}}}}$ Evaluate the limit, using L’Hôpital’s Rule if necessary.

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