Zion Wheeler

Answered

2022-06-25

Evaluate limit of $\underset{x\to 0}{lim}\frac{(1+x{)}^{(1/5)}-(1-x{)}^{(1/5)}}{x}$

Answer & Explanation

pyphekam

Expert

2022-06-26Added 27 answers

Start by adding and subtracting 1 in the numerator:

$\underset{x\to 0}{lim}\frac{(1+x{)}^{\frac{1}{5}}-1+1-(1-x{)}^{\frac{1}{5}}}{x}$

Now split up the limit like this:

$\underset{x\to 0}{lim}\frac{(1+x{)}^{\frac{1}{5}}-1}{x}+\underset{x\to 0}{lim}\frac{(1-x{)}^{\frac{1}{5}}-1}{-x}$

Using the substitution $u=-x$ on the second limit, you should see that both limits are the definition of the derivative of $f(x)={x}^{\frac{1}{5}}$ at $x=1$

${f}^{\prime}({x}_{0})=\underset{h\to 0}{lim}\frac{f({x}_{0}+h)-f({x}_{0})}{h}$

${f}^{\prime}(1)=\underset{h\to 0}{lim}\frac{(1+h{)}^{\frac{1}{5}}-{1}^{\frac{1}{5}}}{h}$

Using the power rule, we know that ${f}^{\prime}(x)=\frac{1}{5}{x}^{-\frac{4}{5}}.$ So, the answer is $2{f}^{\prime}(1)=2(\frac{1}{5}(1{)}^{-\frac{4}{5}})=\overline{){\displaystyle \frac{2}{5}.}}$

Alternatively, you can see this by using the alternative definition ${f}^{\prime}(1)=\underset{x\to 0}{lim}\frac{f(1+x)-f(1-x)}{2x}$ and multiplying the numerator and denominator of our original limit by 2

$\underset{x\to 0}{lim}\frac{(1+x{)}^{\frac{1}{5}}-1+1-(1-x{)}^{\frac{1}{5}}}{x}$

Now split up the limit like this:

$\underset{x\to 0}{lim}\frac{(1+x{)}^{\frac{1}{5}}-1}{x}+\underset{x\to 0}{lim}\frac{(1-x{)}^{\frac{1}{5}}-1}{-x}$

Using the substitution $u=-x$ on the second limit, you should see that both limits are the definition of the derivative of $f(x)={x}^{\frac{1}{5}}$ at $x=1$

${f}^{\prime}({x}_{0})=\underset{h\to 0}{lim}\frac{f({x}_{0}+h)-f({x}_{0})}{h}$

${f}^{\prime}(1)=\underset{h\to 0}{lim}\frac{(1+h{)}^{\frac{1}{5}}-{1}^{\frac{1}{5}}}{h}$

Using the power rule, we know that ${f}^{\prime}(x)=\frac{1}{5}{x}^{-\frac{4}{5}}.$ So, the answer is $2{f}^{\prime}(1)=2(\frac{1}{5}(1{)}^{-\frac{4}{5}})=\overline{){\displaystyle \frac{2}{5}.}}$

Alternatively, you can see this by using the alternative definition ${f}^{\prime}(1)=\underset{x\to 0}{lim}\frac{f(1+x)-f(1-x)}{2x}$ and multiplying the numerator and denominator of our original limit by 2

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