If lim _{x rightarrow 1} frac{sqrt[4]{x^{-3}} | vedclass

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(A) Given $\lim _{x \rightarrow 1} \frac{x^{-3/4} + a x^{5/4}}{x-1} = -2$. For the limit to exist,the numerator must be $0$ at $x=1$,so $1+a=0$,which

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$6$

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(A) Given $\lim _{x ightarrow 1} \frac{x^{-3/4} + a x^{5/4}}{x-1} = -2$. For the limit to exist,the numerator must be $0$ at $x=1$,so $1+a=0$,which gives $a=-1$. Checking the limit with $a=-1$: $\lim _{x ightarrow 1} \frac{x^{-3/4} - x^{5/4}}{x-1} = \lim _{x ightarrow 1} \frac{x^{-3/4}(1 - x^2)}{x-1} = \lim _{x ightarrow 1} x^{-3/4} \frac{(1-x)(1+x)}{-(1-x)} = -2$. Thus,$a=-1$. Now,consider the expansion of $(x^{-3/4} - x^{5/4})^4 = \sum_{k=0}^{4} {^4C_k} (x^{-3/4})^{4-k} (-x^{5/4})^k$. The general term is $T_{k+1} = {^4C_k} (x^{-3/4})^{4-k} (-1)^k (x^{5/4})^k = {^4C_k} (-1)^k x^{-3 + \frac{3k}{4} + \frac{5k}{4}} = {^4C_k} (-1)^k x^{2k-3}$. We want the coefficient of $x^1$,so set $2k-3 = 1$,which gives $2k=4$,so $k=2$. The coefficient is ${^4C_2} (-1)^2 = 6 	imes 1 = 6$.

If $\lim _{x \rightarrow 1} \frac{\sqrt[4]{x^{-3}}+a \sqrt[4]{x^5}}{x-1}=-2$,then the coefficient of $x$ in the expansion of $(\sqrt[4]{x^{-3}}+a \sqrt[4]{x^5})^4$ is

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