1. Find \(lt_{(x,y,z)\rightarrow(0,0,0)}\frac{y^2.z^2}{x^3+x^2.(y)^{\frac{4}{3}}+x^2.(z)^{\frac{4}{3}}}\)
a) 1
b) 0
c) ∞
d) Does Not Exist
Explanation: Put x = t : y = a1 * t3⁄4 : z = a2 * t3⁄4
\(lt_{(x,y,z)\rightarrow(0,0,0)}\frac{(a_1)^2.t^{\frac{3}{2}}.(a_2)^2.t^{\frac{3}{2}}}{t^3+t^2.t.(a_1)^{\frac{4}{3}}+t^2.t.(a_2)^{\frac{4}{3}}}\)
\(lt_{(x,y,z)\rightarrow(0,0,0)}\frac{t^3}{t^3}\times \frac{(a_1)^2.(a_2)^2}{1+(a_1)^{\frac{4}{3}}+(a_2)^{\frac{4}{3}}}\)
\(lt_{(x,y,z)\rightarrow(0,0,0)} \frac{(a_1)^2.(a_2)^2}{1+(a_1)^{\frac{4}{3}}+(a_2)^{\frac{4}{3}}}\)
By varying a1 : a2 one can get different limit values.
2. Find \(lt_{(x,y,z)\rightarrow(0,0,0)}\frac{sin(x).sin(y)}{x.z}\)
a) ∞
b) 1⁄3
c) 1
d) Does Not Exist
Explanation: Put x = t : y = at : z = t
=\(lt_{t\rightarrow 0}\frac{sin(t).sin(at)}{t^2}\)
=\(lt_{t\rightarrow 0}\frac{sin(t)}{t}\times (a) \times lt_{t\rightarrow 0}\frac{sin(at)}{at}\)
= (1) * (a) * (1) = a
3. Find \(lt_{(x,y,z)\rightarrow(2,2,4)}\frac{x^2+y^2-z^2+2xy}{x+y-z}\)
a) ∞
b) 123
c) 9098
d) 8
Explanation: Simplifying the expression yields
\(lt_{(x,y,z)\rightarrow(0,0,0)}\frac{(x+y)^2-z^2}{(x+y)-z}\)
\(lt_{(x,y,z)\rightarrow(0,0,0)}\frac{(x+y+z).(x+y-z)}{(x+y-z)}\)
\(lt_{(x,y,z)\rightarrow(0,0,0)}(x+y+z)=2+2+4\)
=8
4. Find \(lt_{(x,y,z,w)\rightarrow(0,0,0,0)}\frac{x^{-6}.y^2.(z.w)^3}{x+y^2+z-w}\)
a) 1990
b) ∞
c) Does Not Exist
d) 0
Explanation: Put x = t : y = a1.t1⁄2 : z = a2.t : w = a3.t
\(lt_{(x,y,z,w)\rightarrow(0,0,0,0)}\frac{t^{-6}.t.(a_1)^2.t^6.(a_2)^3.(a_3)^3}{t+t.(a_1)^2+a_2.t-a_3.t}\)
\(lt_{(x,y,z,w)\rightarrow(0,0,0,0)}\frac{t}{t}\times \frac{(a_1)^2.(a_2)^3.(a_3)^3}{1+(a_1)^2+a_2-a_3}\)
\(lt_{(x,y,z,w)\rightarrow(0,0,0,0)} \frac{(a_1)^2.(a_2)^3.(a_3)^3}{1+(a_1)^2+a_2-a_3}\)
By changing the values of a1 : a2 : a3 we get different values of limit.
Hence, Does Not Exist is the right answer.
5. Find \(lt_{(x,y,z,w)\rightarrow(3,1,1,11)}\frac{x^4+y^2+z^2+2x^2y+2yz+2x^2z-(w)^2}{x^2+y+z-w}\)
a) 700
b) 701
c) 699
d) 22
Explanation: Simplifying the expression we have
\(lt_{(x,y,z,w)\rightarrow(3,1,1,11)}\frac{(x^2+y+z)^2-(w)^2}{x^2+y+z-w}\)
\(lt_{(x,y,z,w)\rightarrow(3,1,1,11)}\frac{(x^2+y+z+w).(x^2+y+z-w)}{x^2+y+z-w}\)
\(lt_{(x,y,z,w)\rightarrow(3,1,1,11)}(x^2+y+z+w)\)=(32+1+1+11)
=9+1+1+11=22
6. Given that limit exists find \(lt_{(x,y,z)\rightarrow(-2,-2,-2)}\frac{sin((x+2)(y+5)(z+1))}{(x+2)(y+7)}\)
a) 1
b) 3⁄5
c) 1⁄2
d) 0
Explanation: Given that limit exists we can parameterize the curve
Put x = t : y = t : z = t
\(lt_{t\rightarrow -2}\frac{sin((t+2)(t+5)(t+1))}{(t+2)(t+7)}\)
\(lt_{t\rightarrow -2}\frac{sin((t+2)(t+5)(t+1))}{(t+2)(t+5)(t+1)}\times lt_{t\rightarrow -2}\frac{(t+5)(t+1)}{(t+7)}\)
\((1)\times \frac{(-2+5)(-2+1)}{(-2+7)}\)
=\(\frac{(3).(1)}{(5)}=\frac{3}{5}\)
7. Given that limit exist find \(lt_{(x,y,z)\rightarrow(-9,-9,-9)}\frac{tan((x+9)(y+11)(z+7))}{(x+9)(y+10)}\)
a) 2
b) 1
c) 4
d) 3
Explanation: We can parameterize the curve by
x = y = z = t
\(lt_{t\rightarrow -9}\frac{tan((t+9)(t+11)(t+7))}{(t+9)(t+10)}\)
\(lt_{t\rightarrow -9}\frac{tan((t+9)(t+11)(t+7))}{(t+9)(t+11)(t+7)}\times lt_{t\rightarrow -9}\frac{tan((t+11)(t+7))}{t+10}\)
\(=\frac{(-9+11)(-9+7)}{(-9+10)}=\frac{(2)(2)}{(1)}\)
=4
8. Given that limit exists find \(lt_{(x,y,z)\rightarrow(-1,-1,-1)}\frac{tan((x-1)(y-2)(z-3))}{(x-1)(y-6)(z+7)}\)
a) 1
b) 1⁄2
c) 1⁄7
d) 2⁄7
Explanation: We can parameterize the curve by
x=y=z=t
\(lt_{t\rightarrow -1}\frac{tan((t-1)(t-2)(t-3))}{(t-1)(t-6)(t+7)}\)
\(lt_{t\rightarrow -1}\frac{tan((t-1)(t-2)(t-3))}{(t-1)(t-2)(t-3)}\times lt_{t\rightarrow -1}\frac{(t-2)(t-3)}{(t-6)(t+7)} \)
=\(\frac{(-1-2)(-1-3)}{(-1-6)(-1+7)}=\frac{(3)(4)}{(7)(6)}\)
=\(\frac{12}{42}=\frac{2}{7}\)
9. Given that limit exists find \(lt_{(x,y,z)\rightarrow(2,2,2)}\left (\frac{ln(1+\frac{xy-2x-y+z}{xz-2x-6z+12}+\frac{xz-5x-2z+10}{xy-7y-2x+14}}{(x-2)(y-2)(z-2)}\right )\)
a) ∞
b) 1
c) 0
d) ln(4⁄5)
Explanation: We can parameterize the curve by
x = y = z = t
\(lt_{t\rightarrow 2}\left (\frac{ln(1+\frac{xy-2x-y+z}{xz-2x-6z+12}+\frac{xz-5x-2z+10}{xy-7y-2x+14}}{(x-2)(y-2)(z-2)}\right )\)
\(=lt_{t\rightarrow 2}\left (\frac{ln(1+\frac{t}{(t-6)}+\frac{(t-5)}{(t-7)}}{(t-2)^3}\right )\)
\(=lt_{t\rightarrow 2}\left (\frac{ln(1+\frac{2}{(t-6)}+\frac{(2-5)}{(2-7)})}{(2-2)^3}\right )=\frac{ln(\frac{4}{5})}{0}\rightarrow\infty\)
10. Given that limit exists \(lt_{(x,y,z)\rightarrow(0,0,0)}\left (\frac{cos(\frac{\pi}{2}-x).tan(y).cot(\frac{\pi}{2}-z)}{sin(x).sin(y).sin(z)}\right )\)
a) 99
b) 0
c) 1
d) 100
Explanation: Put x = y = z = t
\(lt_{t\rightarrow 0}\left (\frac{cos(\frac{\pi}{2}-t).tan(y).cot(\frac{\pi}{2}-t)}{sin(x).sin(y).sin(z)}\right)\)
\(lt_{t\rightarrow 0}\frac{(sin(t))(tan^2(t))}{sin^3(t)}\)
\(=lt_{t\rightarrow 0}\frac{tan^2(t)}{sin(t)} = lt_{t\rightarrow 0}\frac{1}{cos^2(t)}\)
\(=\frac{1}{cos^2(0)}=\frac{1}{1}=1\)