Question

lim x=1 x^1/3-1/x^1/6-1

Answer 1

2

Answer 2

The given limit is: $\lim_{x \to 1} \frac{x^{1/3} - 1}{x^{1/6} - 1}$

To evaluate this limit, we can use a substitution to simplify the expression.
Let $t = x^{1/6}$.
Then, $t^2 = (x^{1/6})^2 = x^{2/6} = x^{1/3}$.
As $x$ approaches $1$, $t = x^{1/6}$ also approaches $1^{1/6}$, which is $1$.
Substituting these into the limit expression: $\lim_{t \to 1} \frac{t^2 - 1}{t - 1}$ The numerator $t^2 - 1$ is a difference of squares, which can be factored as $(t-1)(t+1)$.
So the expression becomes: $\lim_{t \to 1} \frac{(t-1)(t+1)}{t-1}$ Since $t \to 1$, $t \neq 1$, which means $t-1 \neq 0$. Therefore, we can cancel out the $(t-1)$ term from the numerator and the denominator: $\lim_{t \to 1} (t+1)$ Now, substitute $t=1$ into the simplified expression: $1 + 1 = 2$

The value of the limit is $2$.