- #1
Parmenides
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Let ##L_1## and ##L_2## be two linear mappings from ##R \rightarrow R^n## satisfying the following theorem:
"The mapping ##f: R \rightarrow R^n## is differentiable at ##a \in R## if and only if there exists a linear mapping ##L: R \rightarrow R^n## such that
[tex]\lim_{h \to 0}\frac{f(a + h) - f(a) - L(h)}{h} = {\bf{0}}[/tex]
in which case L is defined by ##L(h) = df_a(h) = hf'(a)##." (From Edwards 'Advanced Calculus').
I am to prove that ##L_1 = L_2##. The text hints that I should first show that:
[tex]lim_{h \to 0}\frac{L_1(h) - L_2(h)}{h} = 0[/tex]
My first idea was to use the definition of ##L## from the theorem so that
[tex]L_1(h) - L_2(h) = h\lim_{h \to 0}\Big(\frac{f(a + h) - f(a)}{h}\Big) - h\lim_{h \to 0}\Big(\frac{f(a + h) - f(a}{h}\Big) = h\lim_{h \to 0}\Big(\frac{f(a + h) - f(a) - f(a + h) + f(a)}{h}\Big) = h\lim_{h \to 0}(0) = 0[/tex]
And so [tex]lim_{h \to 0}\frac{L_1(h) - L_2(h)}{h} = \lim_{h \to 0}\frac{0}{h} = 0[/tex]
But looking back, I think that whole process rested on the assumption that ##L_1 = L_2## in the first place. I am not quite sure what to do.
"The mapping ##f: R \rightarrow R^n## is differentiable at ##a \in R## if and only if there exists a linear mapping ##L: R \rightarrow R^n## such that
[tex]\lim_{h \to 0}\frac{f(a + h) - f(a) - L(h)}{h} = {\bf{0}}[/tex]
in which case L is defined by ##L(h) = df_a(h) = hf'(a)##." (From Edwards 'Advanced Calculus').
I am to prove that ##L_1 = L_2##. The text hints that I should first show that:
[tex]lim_{h \to 0}\frac{L_1(h) - L_2(h)}{h} = 0[/tex]
My first idea was to use the definition of ##L## from the theorem so that
[tex]L_1(h) - L_2(h) = h\lim_{h \to 0}\Big(\frac{f(a + h) - f(a)}{h}\Big) - h\lim_{h \to 0}\Big(\frac{f(a + h) - f(a}{h}\Big) = h\lim_{h \to 0}\Big(\frac{f(a + h) - f(a) - f(a + h) + f(a)}{h}\Big) = h\lim_{h \to 0}(0) = 0[/tex]
And so [tex]lim_{h \to 0}\frac{L_1(h) - L_2(h)}{h} = \lim_{h \to 0}\frac{0}{h} = 0[/tex]
But looking back, I think that whole process rested on the assumption that ##L_1 = L_2## in the first place. I am not quite sure what to do.