Evaluate the limit as x approaches 0 of (1 + tan 5x − cos x) divided by (the square root of 1 − x squared, minus the fifth root of 1 + x).

Step-by-step solution

1. 1

   Verify the 0/0 form. At $x = 0$ the numerator is $1 + \tan 0 - \cos 0 = 1 + 0 - 1 = 0$ and the denominator is $\sqrt{1} - 1 = 0$. Both vanish, so series expansions to the first surviving order will decide the limit.

2. 2

   Expand the numerator. Using $\tan u = u + O(u^3)$ with $u = 5x$, and $\cos x = 1 - \frac{x^2}{2} + O(x^4)$:

   $1 + \tan 5x - \cos x = 1 + 5x - \left(1 - \frac{x^2}{2}\right) + O(x^3) = 5x + \frac{x^2}{2} + O(x^3)$

   The constants $1$ and $-1$ cancel, leaving a leading term of $5x$ — order one.

3. 3

   Expand the denominator with the generalised binomial series. For $(1 + u)^\alpha = 1 + \alpha u + O(u^2)$:

   $\sqrt{1 - x^2} = (1 - x^2)^{1/2} = 1 - \frac{x^2}{2} + O(x^4)$

   $(1 + x)^{1/5} = 1 + \frac{x}{5} + O(x^2)$

   The key asymmetry: the square root has no linear term (its argument is $-x^2$), while the fifth root does.

4. 4

   Subtract to find the leading behaviour of the denominator.

   $\sqrt{1 - x^2} - (1 + x)^{1/5} = \left(1 - \frac{x^2}{2}\right) - \left(1 + \frac{x}{5}\right) + O(x^2) = -\frac{x}{5} + O(x^2)$

   So the denominator is also order one, and negative for small $x > 0$.

5. 5

   Cancel one factor of x and evaluate.

   $\lim_{x \to 0} \frac{x\left(5 + O(x)\right)}{x\left(-\frac{1}{5} + O(x)\right)} = \frac{5}{-\frac{1}{5}} = -25$

6. 6

   Confirm numerically from both sides. At $x = 10^{-5}$ the quotient is $-24.99950$ and at $x = -10^{-5}$ it is $-25.00050$; at $x = \pm 10^{-6}$ the values tighten to $-24.99995$ and $-25.00005$. The two-sided limit is $-25$.