Notes: `lim_(x->5^(+)) sqrt(25-x^2)/(x-5)` Evaluate the limit, using L’Hôpital’s Rule if necessary.

Saturday, 17 May 2014

$\displaystyle\lim_{{{x}\to{{5}}^{{+}}}}\frac{\sqrt{{{25}-{{x}}^{{2}}}}}{{{x}-{5}}}$ Evaluate the limit, using L’Hôpital’s Rule if necessary.

Given to solve,

$\displaystyle\lim_{{{x}\to{{5}}^{{+}}}}\frac{\sqrt{{{25}-{{x}}^{{2}}}}}{{{x}-{5}}}$

Removing the negative form the denominator we get

= $\displaystyle-\lim_{{{x}\to{{5}}^{{+}}}}\frac{\sqrt{{{25}-{{x}}^{{2}}}}}{{{5}-{x}}}$

= $\displaystyle-\lim_{{{x}\to{{5}}^{{+}}}}\frac{\sqrt{{{{5}}^{{2}}-{{x}}^{{2}}}}}{{{5}-{x}}}$

= $\displaystyle-\lim_{{{x}\to{{5}}^{{+}}}}\frac{\sqrt{{{\left({5}-{x}\right)}{\left({5}+{x}\right)}}}}{{{5}-{x}}}$

= $\displaystyle-\lim_{{{x}\to{{5}}^{{+}}}}\sqrt{{\frac{{{5}+{x}}}{{{5}-{x}}}}}$

= $\displaystyle-{\left[\frac{{\lim_{{{x}\to{{5}}^{{+}}}}\sqrt{{{\left({5}+{x}\right)}}}}}{{\lim_{{{x}\to{{5}}^{{+}}}}\sqrt{{{5}-{x}}}}}\right]}$

= $\displaystyle-\frac{\sqrt{{{5}+\lim_{{{x}\to{{5}}^{{+}}}}{x}}}}{\sqrt{{{5}-\lim_{{{x}\to{{5}}^{{+}}}}{x}}}}$

as $\displaystyle{x}\to{{5}}^{{+}}$ ,then the denominator tends from 0 to $\displaystyle-{1}$ .

so,

$\displaystyle\lim_{{{x}\to{{5}}^{{+}}}}\frac{\sqrt{{{25}-{{x}}^{{2}}}}}{{{x}-{5}}}=\infty$

Given to solve,

$\displaystyle\lim_{{{x}\to{{5}}^{{+}}}}\frac{\sqrt{{{25}-{{x}}^{{2}}}}}{{{x}-{5}}}$

Removing the negative form the denominator we get

= $\displaystyle-\lim_{{{x}\to{{5}}^{{+}}}}\frac{\sqrt{{{25}-{{x}}^{{2}}}}}{{{5}-{x}}}$

= $\displaystyle-\lim_{{{x}\to{{5}}^{{+}}}}\frac{\sqrt{{{{5}}^{{2}}-{{x}}^{{2}}}}}{{{5}-{x}}}$

= $\displaystyle-\lim_{{{x}\to{{5}}^{{+}}}}\frac{\sqrt{{{\left({5}-{x}\right)}{\left({5}+{x}\right)}}}}{{{5}-{x}}}$

= $\displaystyle-\lim_{{{x}\to{{5}}^{{+}}}}\sqrt{{\frac{{{5}+{x}}}{{{5}-{x}}}}}$

= $\displaystyle-{\left[\frac{{\lim_{{{x}\to{{5}}^{{+}}}}\sqrt{{{\left({5}+{x}\right)}}}}}{{\lim_{{{x}\to{{5}}^{{+}}}}\sqrt{{{5}-{x}}}}}\right]}$

= $\displaystyle-\frac{\sqrt{{{5}+\lim_{{{x}\to{{5}}^{{+}}}}{x}}}}{\sqrt{{{5}-\lim_{{{x}\to{{5}}^{{+}}}}{x}}}}$

as $\displaystyle{x}\to{{5}}^{{+}}$ ,then the denominator tends from 0 to $\displaystyle-{1}$ .

so,

$\displaystyle\lim_{{{x}\to{{5}}^{{+}}}}\frac{\sqrt{{{25}-{{x}}^{{2}}}}}{{{x}-{5}}}=\infty$

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