Limx - ax5-8-a5-8x1-3-a1-3-

Using lim _x → ax^p a^px^q a^q=pq a^p q lim_x → ax^5/8 a^5/8x^1/3 a^1/3 =5813· a^58 13=158 a^7/24 ...

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Standard XII

Mathematics

Arithmetic Progression

limx → ax5/8-...

Question

$lim x \to a \frac{x^{5 / 8} - a^{5 / 8}}{x^{1 / 3} - a^{1 / 3}} =$

\frac{15}{8} a^{\frac{7}{24}}

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\frac{15}{4} a^{\frac{7}{24}}

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- \frac{15}{8} a^{\frac{7}{24}}

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\frac{15}{4} a^{- \frac{7}{24}}

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Solution

The correct option is A $\frac{15}{8} a^{\frac{7}{24}}$
Using $lim x \to a \frac{x^{p} - a^{p}}{x^{q} - a^{q}} = \frac{p}{q} a^{p - q}$

$lim x \to a \frac{x^{5 / 8} - a^{5 / 8}}{x^{1 / 3} - a^{1 / 3}} = \frac{\frac{5}{8}}{\frac{1}{3}} \cdot a^{\frac{5}{8} - \frac{1}{3}} = \frac{15}{8} a^{7 / 24}$

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Similar questions

lim x \to a \frac{x^{\frac{5}{8}} - a^{\frac{5}{8}}}{x^{\frac{1}{3}} - a^{\frac{1}{3}}} =

lim x \to a \frac{x^{\frac{5}{8}} - a^{\frac{5}{8}}}{x^{\frac{1}{3}} - a^{\frac{1}{3}}} =

lim x \to a {⎧ ⎨ ⎩ ⎡ ⎣ {(\frac{a^{1 / 2} + x^{1 / 2}}{a^{1 / 4} - x^{1 / 4}})}^{- 1} - \frac{2 (a x)^{1 / 4}}{x^{3 / 4} a^{1 / 4} x^{1 / 2} + a^{1 / 2} x^{1 / 4}} - a^{3 / 4} - \sqrt{2^{l o g_{4} a}} ⎤ ⎦ ⎫ ⎬ ⎭}^{8}

a_{1}, a_{2}, a_{3}, a_{4}, a_{5}

are first five terms of an A.P. such that

a_{1} + a_{3} + a_{5} = - 12

and

a_{1} a_{2} a_{3} = 8

. Find the first term and the common difference.

The sum of squares of the intercepts on the coordinate axes made by the tangent to $x^{1 / 3} + y^{1 / 3} = a^{1 / 3} a t (\frac{a}{8}, \frac{a}{8})$ is 2, and a > 0 then a = ____

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limx→ax5/8−a5/8x1/3−a1/3=

The correct option is A 158a724Using limx→axp−apxq−aq=pqap−q limx→ax5/8−a5/8x1/3−a1/3=5813⋅a58−13=158a7/24

$lim x \to a \frac{x^{5 / 8} - a^{5 / 8}}{x^{1 / 3} - a^{1 / 3}} =$

The correct option is A $\frac{15}{8} a^{\frac{7}{24}}$
Using $lim x \to a \frac{x^{p} - a^{p}}{x^{q} - a^{q}} = \frac{p}{q} a^{p - q}$

$lim x \to a \frac{x^{5 / 8} - a^{5 / 8}}{x^{1 / 3} - a^{1 / 3}} = \frac{\frac{5}{8}}{\frac{1}{3}} \cdot a^{\frac{5}{8} - \frac{1}{3}} = \frac{15}{8} a^{7 / 24}$