{"id":32450,"date":"2025-06-22T13:59:52","date_gmt":"2025-06-22T13:59:52","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=32450"},"modified":"2025-06-22T13:59:54","modified_gmt":"2025-06-22T13:59:54","slug":"v2e-x-i5-evaluate-iilm-xm43-vigx-m","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/v2e-x-i5-evaluate-iilm-xm43-vigx-m\/","title":{"rendered":"V2e -x -i5 Evaluate IIlm Xm43 VigX M"},"content":{"rendered":"\n

V2e -x -i5 Evaluate IIlm Xm43 VigX M<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

We are given the following limit expression:lim\u2061x\u2192328\u2212x\u2212519\u2212x\u22124\\lim_{x \\to 3} \\frac{\\sqrt{28 – x} – 5}{\\sqrt{19 – x} – 4}x\u21923lim\u200b19\u2212x\u200b\u2212428\u2212x\u200b\u22125\u200b<\/p>\n\n\n\n

\n\n\n\n

Step 1: Plug in the value of x=3x = 3x=3<\/h3>\n\n\n\n

28\u22123\u2212519\u22123\u22124=25\u2212516\u22124=5\u221254\u22124=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\u22123\u200b\u2212428\u22123\u200b\u22125\u200b=16\u200b\u2212425\u200b\u22125\u200b=4\u221245\u22125\u200b=00\u200b<\/p>\n\n\n\n

This gives us an indeterminate form. So we must simplify.<\/p>\n\n\n\n

\n\n\n\n

Step 2: Use rationalization<\/h3>\n\n\n\n

We can multiply the numerator and denominator by their conjugates to simplify. Start with the numerator:<\/p>\n\n\n\n

Multiply the numerator and denominator by 28\u2212x+5\\sqrt{28 – x} + 528\u2212x\u200b+5:28\u2212x\u2212519\u2212x\u22124\u22c528\u2212x+528\u2212x+5=(28\u2212x)\u221225(19\u2212x\u22124)(28\u2212x+5)=3\u2212x(19\u2212x\u22124)(28\u2212x+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\u2212x\u200b\u2212428\u2212x\u200b\u22125\u200b\u22c528\u2212x\u200b+528\u2212x\u200b+5\u200b=(19\u2212x\u200b\u22124)(28\u2212x\u200b+5)(28\u2212x)\u221225\u200b=(19\u2212x\u200b\u22124)(28\u2212x\u200b+5)3\u2212x\u200b<\/p>\n\n\n\n

Now rationalize the denominator expression 19\u2212x\u22124\\sqrt{19 – x} – 419\u2212x\u200b\u22124 by multiplying top and bottom by its conjugate 19\u2212x+4\\sqrt{19 – x} + 419\u2212x\u200b+4:3\u2212x(19\u2212x\u22124)(28\u2212x+5)\u22c519\u2212x+419\u2212x+4=(3\u2212x)(19\u2212x+4)(19\u2212x\u221216)(28\u2212x+5)=(3\u2212x)(19\u2212x+4)(3\u2212x)(28\u2212x+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\u2212x\u200b\u22124)(28\u2212x\u200b+5)3\u2212x\u200b\u22c519\u2212x\u200b+419\u2212x\u200b+4\u200b=(19\u2212x\u221216)(28\u2212x\u200b+5)(3\u2212x)(19\u2212x\u200b+4)\u200b=(3\u2212x)(28\u2212x\u200b+5)(3\u2212x)(19\u2212x\u200b+4)\u200b<\/p>\n\n\n\n

Now cancel the common factor 3\u2212x3 – x3\u2212x (note: x\u21923x \\to 3x\u21923, but x\u22603x \\ne 3x\ue020=3):19\u2212x+428\u2212x+5\\frac{\\sqrt{19 – x} + 4}{\\sqrt{28 – x} + 5}28\u2212x\u200b+519\u2212x\u200b+4\u200b<\/p>\n\n\n\n

\n\n\n\n

Step 3: Now evaluate the limit as x\u21923x \\to 3x\u21923<\/h3>\n\n\n\n

19\u22123+428\u22123+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\u22123\u200b+519\u22123\u200b+4\u200b=25\u200b+516\u200b+4\u200b=5+54+4\u200b=108\u200b=54\u200b<\/p>\n\n\n\n

\n\n\n\n

Final Answer:<\/h3>\n\n\n\n

45\\boxed{\\frac{4}{5}}54\u200b\u200b<\/p>\n\n\n\n

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.<\/p>\n\n\n\n

\"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

V2e -x -i5 Evaluate IIlm Xm43 VigX M The Correct Answer and Explanation is: We are given the following limit expression:lim\u2061x\u2192328\u2212x\u2212519\u2212x\u22124\\lim_{x \\to 3} \\frac{\\sqrt{28 – x} – 5}{\\sqrt{19 – x} – 4}x\u21923lim\u200b19\u2212x\u200b\u2212428\u2212x\u200b\u22125\u200b Step 1: Plug in the value of x=3x = 3x=3 28\u22123\u2212519\u22123\u22124=25\u2212516\u22124=5\u221254\u22124=00\\frac{\\sqrt{28 – 3} – 5}{\\sqrt{19 – 3} – 4} = \\frac{\\sqrt{25} – 5}{\\sqrt{16} […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-32450","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/32450","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=32450"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/32450\/revisions"}],"predecessor-version":[{"id":32453,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/32450\/revisions\/32453"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=32450"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=32450"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=32450"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}