How can I turn a $0^0$ form into a $\frac{\infty}{\infty}$ or $\frac{0}{0}$ form?

Question

I am trying to evaluate this limit:

$$\lim_{x\to0^{+}}(x-\sin x)^{\frac{1}{\log x}}$$

It's a $0^0$ intedeterminate form, and I am unsure how to deal with it. I have a feeling that if I could turn it to a form where L'Hopital's rule is applicable, then I'd have a chance at solving the problem.

Is there any consistent way of turning a $0^0$ form into a $\frac{\infty}{\infty}$ or $\frac{0}{0}$ form?

If not, how do you deal with this kind of limits?

Answer 1

The general trick is to apply exponential/logarithm to the expression: $$ \lim_{x \to 0^+} (x - \sin{x})^{\frac{1}{\log{x}}} = \lim_{x \to 0^+} e^{\frac{\log(x - \sin{x})}{\log{x}}} = e^{\lim_{x \to 0^+} \frac{\log(x - \sin{x})}{\log{x}}} $$ Now you can apply L'Hopital Rule to the limit in the exponent. I use $=_L$ to denote equality when L'Hopital Rule is applied. \begin{align*} \lim_{x \to 0^+} \frac{\log(x - \sin{x})}{\log{x}} &=_L \lim_{x \to 0^+} \frac{\frac{1 - \cos{x}}{x - \sin{x}}}{\frac{1}{x}} \\ &= \lim_{x \to 0^+} \frac{x - x\cos{x}}{x - \sin{x}} \\ &=_L \lim_{x \to 0^+} \frac{1 - \cos{x} + x\sin{x}}{1 - \cos{x}} \\ &=_L \lim_{x \to 0^+} \frac{2\sin{x} + x\cos{x}}{\sin{x}} \\ &=_L \lim_{x \to 0^+} \frac{3\cos{x} - x\sin{x}}{\cos{x}} \\ &= \frac{3 - 0}{1} \\ &= 3 \end{align*} Thus the desired limit is $e^3$.

Answer 2

Try converting into an indeterminant product using exponentiation by $e$:

$$\lim_{x \to 0^+} (x - \sin x)^{\frac{1}{\log x}} = \lim_{x \to 0^+} e^{\ln(x - \sin x)(\frac{1}{\log x})}$$ and you can go from here using L'Hopital

Answer 3

$$\lim_{x\to0^{+}}(x-\sin x)^{\frac{1}{\ln x}}$$ $$^\lim_{x\to0^{+}}e^{\ln(x-\sin x)^{\frac{1}{\ln x}}}$$

$$e^{\lim_{x\to0^{+}}\dfrac{\ln(x-\sin x)}{\ln(x)}}$$

Applying L Hospital as we have $\dfrac{\infty}{\infty}$ form

$$e^{\lim_{x\to0^{+}}\dfrac{x(1-\cos x)}{x-\sin x}}=1$$

$$e^{\lim_{x\to0^{+}}\dfrac{x^3(1-\cos x)}{x^2(x-\sin x)}}$$

$$\lim_{x\to0^{+}}\dfrac{x-\sin x}{x^3}=\dfrac{1}{6}$$ $$\lim_{x\to0^{+}}\dfrac{1-\cos x}{x^2}=\dfrac{1}{2}$$

After putting these values , you will get final answer as $e^3$

Answer 4

The best option is to take logs. If $L$ is the desired limit then we have $$\log L=\lim_{x\to 0^{+}}\frac{\log(x-\sin x)} {\log x} $$ The expression under limit above can be rewritten as $$3+\dfrac{\log\dfrac{x-\sin x} {x^3}} {\log x} $$ and the numerator here tends to $\log(1/6)$ and denominator tends to $-\infty $ so that the fraction tends to $0$. It follows that $L=e^3$. The limit $$\lim_{x\to 0^{+}}\frac{x-\sin x} {x^3}=\frac{1}{6}$$ is easily handled by L'Hospital's Rule or Taylor series.