Question

Calculate each limit.

Answer 1

`\sf \lim_(x\to 0) (\tan\left(2x^(2)\right)+\sin^(2)(5x))/(x^(2)) = 27`

Explanation:

let's go through the steps to find the limit using l'Hôpital's rule:

L'Hôpital's Rule is a mathematical technique used for evaluating indeterminate forms, particularly when dealing with limits. It states that if the limit of the ratio of two functions
`\sf (f(x))/(g(x))` as
`\sf x` approaches a certain value is of the form
`\sf (0)/(0)` or
`\sf (\infty)/(\infty)`, then the limit of the original expression is the same as the limit of the ratio of their derivatives:

`\sf \lim_(x \to a) (f(x))/(g(x)) = \lim_(x \to a) (f'(x))/(g'(x))`

This rule is particularly useful when dealing with limits involving functions that approach zero or infinity, and it helps simplify the evaluation of such limits.

Given limit:

`\sf \lim_(x\to 0) (2x^(2)+\sin^(2)(5x))/(x^(2))`

Rearrange terms:

`\sf \lim_(x\to 0) \left((\sin^(2)(5x)+\tan\left(2x^(2)\right))/(x^(2))\right)`

Direct substitution:

`\sf (\sin^(2)(5 \cdot 0)+\tan\left(2 \cdot 0^(2)\right))/(0^(2))`

Simplify:

`\sf (0)/(0)`

Apply l'Hôpital rule:

`\sf \lim_(x\to 0) (10\sin\left(5x\right)\cos\left(5x\right)+4\sec^(2)\left(2x^(2)\right)x)/(2x)`

Direct substitution:

`\sf (10\sin\left(5 \cdot 0\right)\cos\left(5 \cdot 0\right)+4\sec^(2)\left(2 \cdot 0^(2)\right) \cdot 0)/(2 \cdot 0)`

Simplify:

`\sf (0)/(0)`

Apply l'Hôpital's rule again:

`\sf \lim_(x\to 0) (-50\sin^(2)(5x)+50\cos^(2)(5x)+4\sec^(2)\left(2x^(2)\right)+32x^(2)\sec^(2)\left(2x^(2)\right)\tan\left(2x^(2)\right))/(2)`

Direct substitution:

`\sf (-50\sin^(2)(5 \cdot 0)+50\cos^(2)(5 \cdot 0)+4\sec^(2)\left(2 \cdot 0^(2)\right)+32 \cdot 0^(2)\sec^(2)\left(2 \cdot 0^(2)\right)\tan\left(2 \cdot 0^(2)\right))/(2)`

Simplify:

`\sf 27`

So, the answer is:

`\sf \lim_(x\to 0) (\tan\left(2x^(2)\right)+\sin^(2)(5x))/(x^(2)) = 27`