Question

What is the limit? `lim_(theta->0)(tan(3theta^2)+sin^2 5theta)/(theta^2)` and `lim_(x->oo)(cos(pi/x))/(x-2)`

Answer 1

`lim_(theta rarr 0)(tan(3 theta^2)+sin^2 5 theta)/(theta^2) = 28`

`lim_(x rarr oo)cos(pi/x)/(x-2) = 0`

Explanation:

We can start by using some standard, well defined elementary calculus limits. (all trig function evaluated in radians):

# {: (ul("Limit"), ul("Expression"), ul("Result"), ul("Notes")), (A, lim_(theta rarr 0) A+B, =lim_(theta rarr 0) A+lim_(theta rarr 0) B,), (B, lim_(theta rarr 0) AB, =lim_(theta rarr 0) A * lim_(theta rarr 0) B,), (C, lim_(theta rarr 0) sin theta, =0,"Direct Evaluation"), (D, lim_(theta rarr 0) cos theta, =1,"Direct Evaluation"), (E, lim_(theta rarr 0) (sin theta)/theta, =1,"Should be learnt"), (F, lim_(theta rarr 0) (1-cos theta)/theta, =0,"Should be learnt"), (G, lim_(theta rarr 0) (tan theta)/theta, =1,) :} #

Limit 1:

We seek:

`L_1 = lim_(theta rarr 0)(tan(3 theta^2)+sin^2 5 theta)/(theta^2)`

We first note that we cannot evaluate the limit directly as we get an indeterminate form `0/0`. We could use L'Hôpital's rule, but let us evaluate the limit using algebra alone. Although not obvious we can divide each term by the denominator and then manipulate the terms until they match our standard limits, thus:

`L_1 = lim_(theta->0) {tan(3 theta^2)/(theta^2)+sin^2 (5 theta)/(theta^2)}`

`\ \ \ \ = lim_(theta rarr 0) {tan(3 theta^2)/(theta^2) * 3/3 + sin (5 theta)/(theta) * sin (5 theta)/(theta) * 5/5 * 5/5}`

`\ \ \ \ = lim_(theta rarr 0) {(3 tan(3 theta^2))/(3 theta^2) + (5 sin (5 theta))/(5 theta) * (5 sin (5 theta))/(5 theta) }`

Using `[A]` and `[B]`

`L_1 = 3 lim_(theta rarr 0) (tan(3 theta^2))/(3 theta^2) + 5 lim_(theta rarr 0) (sin (5 theta))/(5 theta) * 5 lim_(theta rarr 0) (sin (5 theta))/(5 theta)`

And each remaining limit is in the standard form required by `E` and `G` (this can be demonstrated via a substitution as `theta rarr 0 => 3 theta^2 rarr 0`, and `theta rarr 0 => 5 theta rarr 0`, and so we get:

`L_1 = 3 * 1 + 5 * 1 * 5 * 1`
`\ \ \ \ = 3 + 25`
`\ \ \ \ = 28`

Limit 2:

We seek:

`L_2 = lim_(x rarr oo)cos(pi/x)/(x-2)`

We can manipulate the limit, in preparation for a substitution:

`L_2 = lim_(x rarr oo)cos(pi/x)/(x-2) * (1/x)/(1/x)`
`\ \ \ \ = lim_(x rarr oo) (1/x cos(pi * 1/x)) /(1-2/x)`

Let us now perform a substitution `u=1/x` then we note that as `xrarr 00 => u rarr 0`, and so we can rewrite the limit:

`L_2 = lim_(u rarr 0) (u cos(pi * u)) /(1-u)`
`\ \ \ \ = lim_(u rarr 0) cos(pi u) * u /(1-u)`
`\ \ \ \ = lim_(u rarr 0) cos(pi u) * lim_(u rarr 0) u /(1-u)`

Both of these limits can be evaluated by direct substitution, thus:

`L_2 = cos(0) * 0 /(1-0)`
`\ \ \ \ = 0`

Note 1:

The tangent limit. `E`, can be readily derived:

`lim_(theta rarr 0) (tan theta)/theta = lim_(theta rarr 0) (sin theta)/(cos theta) * 1/theta`

`" " = lim_(theta rarr 0) (sin theta)/(cos theta) * 1/theta`

`" " = lim_(theta rarr 0) 1/(cos theta) * (sin theta)/theta`

`" " = 1 * 1`