Find the limit $\lim_{x\to 0^-}| \left( 1+x^{3} \right)^{1/2}-1-x^{5} |/(\sin x-x)$

Question

I am studying for my first calculus exam (well, it's half an exam), and of course we have to solve limits, without using L'Hospital rule, and using asymptotic analysis.

I can't solve this one $$\lim_{x\rightarrow 0^{-}}\frac{\left| \left( 1+x^{3} \right)^{\frac{1}{2}}-1-x^{5} \right|}{\sin x-x}$$

Can you guide me through the solution? Thanks a lot.

Answer 1

Look at $f(x) =\frac{ ( 1+x^{3} )^{\frac{1}{2}}-1-x^{5} }{\sin x-x} $.

By "asymptotic analysis", as $x \to 0$, $(1+x^3)^{\frac12} \approx 1+\frac12 x^3 + \frac{\frac12 (-\frac12)}{2}(x^3)^2 = 1+\frac12 x^3 - \frac18 x^6 $ and $\sin(x) \approx x-\frac{x^3}{6}+\frac{x^5}{120} $ so $\sin(x)-x \approx -\frac{x^3}{6}+\frac{x^5}{120} $.

Therefore, using the fact that $ax^n+bx^m \approx ax^n $ as $x \to 0$ if $1 \le n < m$ and $a\ne 0$, $f(x) \approx \frac{(1+\frac12 x^3- \frac18 x^6)-1-x^5}{-\frac{x^3}{6}+\frac{x^5}{120}} =\frac{\frac12 x^3 - \frac18 x^6-x^5}{-\frac{x^3}{6}+\frac{x^5}{120}} \approx \frac{\frac12 x^3}{-\frac{x^3}{6}} =-3 $.

Answer 2

Hint: Divide and multiply by $x^3$, and above multiply and divide by the conjugate expression which is $(1+x^3)^{1/2}+(1+x^5)^2$ NExt to compute the fraction below use De l'Hospital rule two times to find $-1/6$ and above after some simplification the result follows easly.