Taylor series of a division-by-zero equation

Question

I need to calculate taylor series of $(\frac1{t^3}+\frac3{t^2})^{1/3} - \sqrt{(\frac1{t^2}-\frac2{t})}$ at $t = 0$

to calculate limit $(\frac1{t^3}+\frac3{t^2})^{1/3} - \sqrt{(\frac1{t^2}-\frac2{t})}$ as $t \rightarrow 0$

I got division-by-zero error where $t = 0$. however, another algebra tool such as wolframalpha and symbolab give me an answer. (Please take a look to the below link)

http://www.wolframalpha.com/input/?i=taylor+%28%281%2Ft%5E3%29%2B3%2Ft%5E2%29%5E%281%2F3%29+-+%281%2Ft%5E2-2%2Ft%29%5E%281%2F2%29+at+t+%3D+0

Does anyone how to get the result ?

Thanks for reading the question.

Answer 1

For positive $t$, our expression is $$\frac{1}{t}(1+3t)^{1/3}-\frac{1}{t}(1-2t)^{1/2}.$$ The limit of this as $t\to 0^+$ is easy to compute using L'Hospital's Rule. For note that the derivative of $(1-3t)^{1/3}-(1-2t)^{1/2}$ is continuous at $t=0$ and has limit $2$. In principle we need to do a separate calculation for $t\to 0^-$.

To get the full Taylor series, find the series for $(1+3t)^{1/3}$, for $(1-2t)^{1/2}$, subtract, and divide term by term by $t$. This can be done, since the constant terms in the two expansions cancel.

Answer 2

Remember that when $x$ is small compared to $1$, $(1+x)^n \simeq (1+n~x)$. So $$(1+3t)^{1/3} \simeq 1+t$$ $$(1-2t)^{1/2} \simeq 1-t$$ and then $$\frac{1}{t}(1+3t)^{1/3}-\frac{1}{t}(1-2t)^{1/2} \simeq \frac{1}{t} (1+t)-\frac{1}{t} (1-t)=2$$.

If you have needed to go further, you could have used the binomial expansion of $$(1+x)^n=1+nx+ \frac {n(n-1)}{2!}x^2+...$$ Applied to your expressions, $$(1+3t)^{1/3} \simeq 1 + t - t^2$$ $$(1-2t)^{1/2} \simeq 1-t-\frac{t^2}{2}$$ $$\frac{1}{t}(1+3t)^{1/3}-\frac{1}{t}(1-2t)^{1/2} \simeq \frac{1}{t} (1+t-t^2)-\frac{1}{t} (1-t-\frac{t^2}{2})=2-\frac{t}{2}$$.