limx→0amx−bnxx
limx→0(amx−1)−(bnx−1)x
=limx→0mamx−1mx−limx→0nbnx−1nx
=m log a−n log b
=log(ambn)
limx→0amx−bnxsin kx
Let f(x)=∣∣ ∣ ∣∣a(x)b(x)c(x)m(x)n(x)l(x)g(x)h(x)k(x)∣∣ ∣ ∣∣ then ∫a0f(x) is∣∣ ∣ ∣∣∫a0a(x)∫a0b(x)∫a0c(x)∫a0m(x)∫a0n(x)∫a0l(x)∫a0g(x)∫a0h(x)∫a0k(x)∣∣ ∣ ∣∣
Find:
(i)limx→2[x]
(ii)limx→52[x]
(iii)limx→1[x]
Evaluate the following one sided limits:
(i)limx→2+x−3x2−4
(ii)limx→2−x−3x2−4
(iii)limx→0+13x
(iv)limx→8+2xx+8
(v)limx→0+2x15
(vi)limx→π−2tan x
(vii)limx→π2+sec x
(viii)limx→0−x2−3x+2x3−2x2
(ix)limx→−2+x2−12x+4
(x)limx→0+(2−cot x)
(xi)limx→0−1+cosecx