1. Find \(lt_{x\rightarrow -5}(\frac{tan^{-1}(x^2+6x+5)}{x^2+15x+50})\)
a) 0
b) 1
c) -4⁄5
d) -1
Explanation: \(=lt_{x\rightarrow -5}\frac{tan^{-1}((x+5)(x+1))}{(x+5)(x+10)}\)
Now expand into Taylor Series for tan-1(x)
\(=lt_{x\rightarrow -5}\frac{1}{(x+5)(x+10)}\times((x+5)(x+1)-\frac{(x+5)^3(x+2)^3}{3}+\frac{(x+5)^5(x+2)^5}{5}-..\infty)\)
\(=lt_{x\rightarrow -5}(\frac{(x+1)}{(x+10)}-\frac{(x+5)^2(x+2)^3}{3(x+10)}+\frac{(x+5)^4(x+2)^5}{5(x+10)}-…\infty)\)
\(=lt_{x\rightarrow -5}\frac{(-5)+1}{(-5)+10}\)
=\(-\frac{4}{5}\)
2. Find \(lt_{x\rightarrow 0}\frac{sin((4x^3)tan^{-1}(x))}{x^4}\)
a) 1
b) 2
c) 4
d) 3
Explanation: Expand into Mclaurin series
\(=lt_{x\rightarrow 0}(\frac{1}{x^4})\times(\frac{4x^3 tan^{-1}(x)}{1!}-\frac{16x^6 tan^{-1}(x)}{3!}+…\infty)\)
\(=lt_{x\rightarrow 0}(\frac{4 tan^{-1}(x)}{x1!}-\frac{16x^2 tan^{-1}(x)}{3!}+…\infty)\)
Neglecting higher order terms (which go to zero) we have
\(=lt_{x\rightarrow 0}(\frac{4 tan^{-1}(x)}{x!})=lt_{x\rightarrow 0}(\frac{4}{x})\times(\frac{x}{1}-\frac{x^3}{3}+..\infty)\)
\(=lt_{x\rightarrow 0}(\frac{4 tan^{-1}(x)}{x!})=lt_{x\rightarrow 0}(\frac{4}{1}-\frac{4x^2}{3}+…\infty)=4\)
3. Find \(lt_{x\rightarrow 0}\frac{sin(sin(x))}{x}\)
a) 1
b) ∞
c) 0
d) -1
Explanation: \(=lt_{x\rightarrow 0}(\frac{1}{x})\times(\frac{sin(x)}{1!}-\frac{(sin(x))^3}{3!}+…\infty)\)
\(=lt_{x\rightarrow 0}(\frac{sin(x)}{x.1!}-\frac{(sin(x))^3}{x.3!}+…\infty)\)
\(=lt_{x\rightarrow 0}(\frac{sin(x)}{x}-lt_{x\rightarrow 0}\frac{sin(x)}{x.3!}\times(sin(x))^2+…\infty)\)
\(=lt_{x\rightarrow 0}\frac{sin(x)}{x}=1\)
4. Find \(=lt_{x\rightarrow 0}\frac{(a_nx^n+a_{n-1}x^{n-1}+…+a_1x+a_0)}{(b_nx^n+b_{n-1}x^{n-1}+…+b_1x+b_0)}\)
a) an ⁄ bn
b) ∞
c) No general form
d) bn ⁄ an
Explanation: \(=lt_{x\rightarrow 0}(\frac{x^n}{x^n})\times\frac{(a_n+a_{n-1}\frac{1}{x}+…+a_1\frac{1}{x^{n-1}}+a_0\frac{1}{x^{n}})}{(b_n+b_{n-1}\frac{1}{x}+…+b_1\frac{1}{x^{n-1}}+b_0\frac{1}{x^{n}})}\)
\(=\frac{a_n}{b_n}\)
5. Find \(lt_{x\rightarrow -2}\frac{sin(\frac{1+(\frac{(x+2)^2(x^2+1)}{x^3+3})}{x+2})}{(x+2)}\)
a) ∞
b) 0
c) 2
d) -∞
Explanation: First evaluate
\(=lt_{x\rightarrow -2}\frac{ln(1+\frac{(x+2)^2(x^2+1)}{x^3+3})}{x+2}\)
\(=lt_{x\rightarrow -2}(\frac{1}{x+2})\times(\frac{(x+2)^2(x^2+1)}{(x^3+3)}-\frac{(x+2)^3(x^3+1)^2}{2.(x^3+3)^2}+…\infty)\)
\(=lt_{x\rightarrow -2}\times(\frac{(x+2)^2(x^2+1)}{(x^3+3)}-\frac{(x+2)^3(x^3+1)^2}{2.(x^3+3)^2}+…\infty)\)
We hence the form for the totle limit as
\(lt_{x\rightarrow a}\frac{sin(f(x))}{g(x)}\)=1
Where f(x)→0 : g(x)→0 as x→a
This is true for the above problem
Thus, we can deduce the limit as
= 1
Hence, 2 is the right answer.
6. Find \(lt_{x\rightarrow 0}\frac{(3e^x-2e^{2x}-e^{3x})}{(e^x+e^{2x}-2e^{3x})}\)
a) 3⁄2
b) 0
c) 4⁄3
d) –4⁄3
Explanation: Form is 0 / 0
Applying L hospitals rule we have
\(=lt_{x\rightarrow 0}\frac{3e^x-4e^{2x}-3e^{3x}}{e^x+2e^{2x}-6e^{3x}}\)
\(=\frac{3-4-3}{1+2-6}\)
\(=\frac{4}{3}\)
7. Find relation between a and b such that the following limit is got after a single application of L hospitals Rule \(lt_{x\rightarrow 0}\frac{ae^x+be^{2x}}{be^x+ae^{2x}}\)
a) b⁄a = 2
b) a⁄b = 2
c) a = b
d) a = -b
Explanation: Given differentiation is applied once we get
ae0 + be0 = 0 = a + b (numerator → zero)
be0 + ae0 = 0 = a + b (denominator → zero)
Thus the relation between (a, b) and is
a + b = 0
OR
a = -b.
8. Find \(lt_{x\rightarrow 0}\frac{2cos(2x)+3cos(5x)-5cos(19x)}{cos(4x)-cos(3x)}\)
a) -76
b) -6
c) -7
d) 0
Explanation: Form here 0⁄0
Applying L hospitals rule we have
\(=lt_{x\rightarrow 0}\frac{4cos(2x)+15cos(5x)-95cos(19x)}{4cos(4x)-3cos(3x)}\)
\(\frac{4+15-95}{4-3}\) = -76.
9. Find how many rounds of differentiation are required to have finite limit for \(lt_{x\rightarrow 0}\frac{cos(ax)+cos(bx)-2cos(cx)}{cos(ax)+2cos(bx)-3cos(cx)}\) given that a ≠ b ≠ c
a) 3
b) 0
c) 2
d) 4
Explanation: Applying L hospitals rule
\(=lt_{x\rightarrow 0}\frac{a.cos(ax)+b.cos(bx)-2c.cos(cx)}{a.cos(ax)+2b.cos(bx)-3c.cos(cx)}=lt_{x\rightarrow 0}\frac{a+b-2c}{a+2b-3c}\)
Assume now that
a + b + 2c = 0 and a + 2b – 3c = 0
We must have
a = c = b but given a ≠ c ≠ b
Thus, our assumption is false and a finite limit exists after first round of differentiation.
Hence, 2 is the right answer.
10. Find \(lt_{p\rightarrow\infty}\frac{p^5.p!}{5.6…(5+p)}\)
a) 4!
b) 5!
c) 0
d) ∞
Explanation: \(=lt_{p\rightarrow\infty}\frac{1.2.3.4}{1.2.3.4}\times \frac{p^5.p!}{5.6…(5+p)}\)
\(=lt_{p\rightarrow\infty}\frac{4!.p^5.p!}{(p+5)!}\)
\(=lt_{p\rightarrow\infty}\frac{4!.p^5}{(p+5)(p+4)(p+3)(p+2)(p+1)}\)
\(=lt_{p\rightarrow\infty}(4!)\times\frac{p^5}{(p+5)(p+4)(p+3)(p+2)(p+1)}=4!\times(1)\)
= 4!