Question

Evaluate the limit with either L'Hôpital's rule or previously learned methods.lim Sin(x)- Tan(x)/ x^3x → 0

Answer 1

`(-1)/(6)`

Explanation:

Given the limit of a function expressed as
`\lim_( x\to \ 0) (sin(x)-tan(x))/(x^3)`, to evaluate the following steps must be carried out.

Step 1: substitute x = 0 into the function

`= (sin(0)-tan(0))/(0^3)\\= (0)/(0) (indeterminate)`

Step 2: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the function

`= \lim_( x\to \ 0) ((d)/(dx)[ sin(x)-tan(x)])/((d)/(dx) (x^3))\\= \lim_( x\to \ 0) (cos(x)-sec^2(x))/(3x^2)\\`

Step 3: substitute x = 0 into the resulting function

`= (cos(0)-sec^2(0))/(3(0)^2)\\= (1-1)/(0)\\= (0)/(0) (ind)`

Step 4: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 2

`= \lim_( x\to \ 0) ((d)/(dx)[ cos(x)-sec^2(x)])/((d)/(dx) (3x^2))\\= \lim_( x\to \ 0) (-sin(x)-2sec^2(x)tan(x))/(6x)\\`

`= (-sin(0)-2sec^2(0)tan(0))/(6(0))\\= (0)/(0) (ind)`

Step 6: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 4

`= \lim_( x\to \ 0) ((d)/(dx)[ -sin(x)-2sec^2(x)tan(x)])/((d)/(dx) (6x))\\= \lim_( x\to \ 0) ([ -cos(x)-2(sec^2(x)sec^2(x)+2sec^2(x)tan(x)tan(x)])/(6)\\\\= \lim_( x\to \ 0) ([ -cos(x)-2(sec^4(x)+2sec^2(x)tan^2(x)])/(6)\\`

Step 7: substitute x = 0 into the resulting function in step 6

`= ([ -cos(0)-2(sec^4(0)+2sec^2(0)tan^2(0)])/(6)\\\\= (-1-2(0))/(6) \\= (-1)/(6)`

Hence the limit of the function
`\lim_( x\to \ 0) (sin(x)-tan(x))/(x^3) \ is \ (-1)/(6)`.