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Question Number 25026 by NECx last updated on 02/Dec/17

$$\underset{x\to 1}{\lim }\left(\frac{x^{1/3}-1}{x^{1/4}-1}\right)$$$$Evaluatethis$$$$$$

Answered by nnnavendu last updated on 02/Dec/17

$$\underset{x\to 1}{\lim }\left(\frac{x^{1/3}-1}{x^{1/4}-1}\right)$$$$\underset{x\to 1}{\lim }\left(\frac{{\left(x^{1/4}\right)}^{4/3}-1^{4/3}}{x^{1/4}-1}\right)$$$$\underset{x\to 1}{\lim }\left(4/31^{\frac{4}{3}-1}\right)$$$$4/3$$$$$$$$f\mathrm{armula}-$$$$\underset{x\to \mathrm{n}}{\lim }\left(\frac{x^{\mathrm{n}}-{\mathrm{a}}^{\mathrm{n}}}{x-\mathrm{a}}\right)={na}^{n-1}$$$$$$

Commented by moxhix last updated on 02/Dec/17

$$Ithink...$$$$\underset{x\to a}{lim}\left(\frac{x^n-a^n}{x-a}\right)={\left(\frac{d}{dx}{x}^n\right)}_{x=a}={na}^{n-1}$$

Answered by jota+ last updated on 03/Dec/17

$$=\frac{\underset{x\to 1}{\lim }{x}^{-2/3}/3}{\underset{x\to 1}{\lim }{x}^{-3/4}/4}=\frac{4}{3}$$$$UseHopitalrule$$$$$$

Commented by NECx last updated on 02/Dec/17

$$thanksbosses....Ireallyappreciatethis$$

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