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#

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 #
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

The limit of (tan(3theta^2)+sin^2 5theta)/(theta^2) as theta approaches 0 is 15.

The limit of (cos(pi/x))/(x-2) as x approaches infinity is 0.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7