In(e-) if x 23 A) lim f (x) doesn't exist B) lim f(x)=0 C) lim f(x)=6 D) limf(x)= 3 E) lim f(x)=-3

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[3] 10. Prove the theorem that states: If f is differentiable at a, then f is continuous at a. Solution: Assume lim (*)={@) = f'(a). Then, since lim (1 – a) = 0, we get by the Limit Laws that f(x) - - @ -a) = lim + f(x)-f(a) T-a lim( 1a -a) = f'(a)-0 = 0 ma a limf (x) - f(a)] = lim- rta Now, observe that lim f(1) = lim[f (2) - f(a) + f(a)] = lim(f(x) – f(a)] + lim f(a) = 0 + f(a) = f(a) Hence, f(x) is continuous at a.

5. If f(x)= -2r* +3x+x-1 then 2* +38° + x? +5x+1 A) lim f(x)=-1 B) lim f(x)=0 C) lim f(x)=-2 E) none of these D) lim f(x) = -6

5. If f(x)= 27° -5x + 3x –7 then A) lim f(n= f(x)=0 B) lim f(x)=0 D) lim f(x)=-- C) lim E) None of these

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In(e-) if x 23 A) lim f (x) doesn't exist B) lim f(x)=0 C) lim f(x)=6 D) limf(x)= 3 E) lim f(x)=-3

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In(e*-*) if x 23 A) lim f (x) doesn't exist B) lim f(x)=0 C) lim f(x)=6 D) limf(x)= 3 E) lim f(x)=-3

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In(e-) if x 23 A) lim f (x) doesn't exist B) lim f(x)=0 C) lim f(x)=6 D) limf(x)= 3 E) lim f(x)=-3