Question

Evaluate the limit? : `lim_(x rarr 0) ( tanx-x ) / (x-sinx)`

Answer 1

`lim_(x rarr 0) ( tanx-x ) / (x-sinx) = 2`

Explanation:

We want to find:

`L = lim_(x rarr 0) ( tanx-x ) / (x-sinx)`

Method 1 : Graphically
graph{( tanx-x ) / (x-sinx) [-8.594, 9.18, -1.39, 7.494]}

Although far from conclusive, it appears that:

`L = 2`

Method 2 : L'Hôpital's rule

The limit is of an indeterminate form `0/0`, and so we can apply L'Hôpital's rule which states that for an indeterminate limit then, providing the limits exists then:

`lim_(x rarr a) f(x)/g(x) = lim_(x rarr a) (f'(x))/(g'(x))`

And so applying L'Hôpital's rule we get:

`L = lim_(x rarr 0) (d/dx ( tanx-x )) / (d/dx (x-sinx))`

`\ \ \ = lim_(x rarr 0) ( sec^2x-1 ) / (1-cosx)`

This limits is also of an indeterminate form, so we can apply L'Hôpital's again to get:

`L = lim_(x rarr 0) (d/dx ( sec^2x-1 ) ) / (d/dx (1-cosx) )`

`\ \ \ = lim_(x rarr 0) ( 2secx(secxtanx) ) / (sinx)`
`\ \ \ = lim_(x rarr 0) ( 2sec^2x(sinx/cosx) ) / (sinx)`
`\ \ \ = lim_(x rarr 0) 2sec^2x(1/cosx)`
`\ \ \ = 2lim_(x rarr 0) sec^3x`
`\ \ \ = 2`