V2e -x -i5 Evaluate IIlm Xm43 VigX M
The Correct Answer and Explanation is:
We are given the following limit expression:limx→328−x−519−x−4\lim_{x \to 3} \frac{\sqrt{28 – x} – 5}{\sqrt{19 – x} – 4}x→3lim19−x−428−x−5
Step 1: Plug in the value of x=3x = 3x=3
28−3−519−3−4=25−516−4=5−54−4=00\frac{\sqrt{28 – 3} – 5}{\sqrt{19 – 3} – 4} = \frac{\sqrt{25} – 5}{\sqrt{16} – 4} = \frac{5 – 5}{4 – 4} = \frac{0}{0}19−3−428−3−5=16−425−5=4−45−5=00
This gives us an indeterminate form. So we must simplify.
Step 2: Use rationalization
We can multiply the numerator and denominator by their conjugates to simplify. Start with the numerator:
Multiply the numerator and denominator by 28−x+5\sqrt{28 – x} + 528−x+5:28−x−519−x−4⋅28−x+528−x+5=(28−x)−25(19−x−4)(28−x+5)=3−x(19−x−4)(28−x+5)\frac{\sqrt{28 – x} – 5}{\sqrt{19 – x} – 4} \cdot \frac{\sqrt{28 – x} + 5}{\sqrt{28 – x} + 5} = \frac{(28 – x) – 25}{(\sqrt{19 – x} – 4)(\sqrt{28 – x} + 5)} = \frac{3 – x}{(\sqrt{19 – x} – 4)(\sqrt{28 – x} + 5)}19−x−428−x−5⋅28−x+528−x+5=(19−x−4)(28−x+5)(28−x)−25=(19−x−4)(28−x+5)3−x
Now rationalize the denominator expression 19−x−4\sqrt{19 – x} – 419−x−4 by multiplying top and bottom by its conjugate 19−x+4\sqrt{19 – x} + 419−x+4:3−x(19−x−4)(28−x+5)⋅19−x+419−x+4=(3−x)(19−x+4)(19−x−16)(28−x+5)=(3−x)(19−x+4)(3−x)(28−x+5)\frac{3 – x}{(\sqrt{19 – x} – 4)(\sqrt{28 – x} + 5)} \cdot \frac{\sqrt{19 – x} + 4}{\sqrt{19 – x} + 4} = \frac{(3 – x)(\sqrt{19 – x} + 4)}{(19 – x – 16)(\sqrt{28 – x} + 5)} = \frac{(3 – x)(\sqrt{19 – x} + 4)}{(3 – x)(\sqrt{28 – x} + 5)}(19−x−4)(28−x+5)3−x⋅19−x+419−x+4=(19−x−16)(28−x+5)(3−x)(19−x+4)=(3−x)(28−x+5)(3−x)(19−x+4)
Now cancel the common factor 3−x3 – x3−x (note: x→3x \to 3x→3, but x≠3x \ne 3x=3):19−x+428−x+5\frac{\sqrt{19 – x} + 4}{\sqrt{28 – x} + 5}28−x+519−x+4
Step 3: Now evaluate the limit as x→3x \to 3x→3
19−3+428−3+5=16+425+5=4+45+5=810=45\frac{\sqrt{19 – 3} + 4}{\sqrt{28 – 3} + 5} = \frac{\sqrt{16} + 4}{\sqrt{25} + 5} = \frac{4 + 4}{5 + 5} = \frac{8}{10} = \frac{4}{5}28−3+519−3+4=25+516+4=5+54+4=108=54
Final Answer:
45\boxed{\frac{4}{5}}54
This result was reached by rationalizing the expressions and carefully simplifying the limit. The initial indeterminate form indicates that we need algebraic manipulation before directly substituting the value.
