Chapter 1: Limits
Section 1.3: Limit Laws
Example 1.3.3
Use the laws in Table 1.3.1 to evaluate limt→−2t2+5tt10−6t5+2tt4+4t2+4.
Solution
This limit yields to the Quotient rule, provided it can be shown that the limits of the numerator and denominator exist, and the limit of the denominator is not zero. The limit of the numerator yields to the product rule, so the limits of the separate factors have to be shown to exist. Hence, the calculations in Table 1.3.3(a).
limt→−2(t2+5t)=−6
The limit of the first factor in the numerator exists.
limt→−2(t10−6t5+2t)=1212
The limit of the second factor in the numerator exists.
limt→−2(t4+4t2+4)=36
The limit of the denominator exists and is not zero.
limt→−2((t2+5t)(t10−6t5+2t)) = −6⋅1212
The limit of the numerator is the product of the limits of the separate factors.
limt→−2(t2+5t)(t10−6t5+2t)t4+4t2+4 = −6⋅121236 = −202
The limit of the given rational function is the quotient of the limits because all appropriate limits exist, and the limit of the denominator is not zero.
Table 1.3.3(a) Quotient rule for limits applied to limt→−2t2+5tt10−6t5+2tt4+4t2+4
Annotated stepwise solution via the Context Panel
Tools≻Load Package: Student Calculus 1
Calculus palette: Limit template
Context Panel: Student Calculus1≻All Solution Steps
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limt→−2t2+5tt10−6t5+2tt4+4t2+4→show solution stepsLimit Stepslimt→−2t2+5tt10−6t5+2tt4+4t2+4▫1. Apply the quotient rule◦Recall the definition of the quotient rulelimt→aftgt=limt→aftlimt→agtf1t=t2+5tf2t=t10−6t5+2tf3t=1t4+4t2+4This gives:limt→−2t2+5tt10−6t5+2tlimt→−2t4+4t2+4▫2. Apply the sum rule◦Recall the definition of the sum rulelimt→af1t+f2t+f3t=limt→af1t+limt→af2t+limt→af3tf1t=t4f2t=4t2f3t=4This gives:limt→−2t2+5tt10−6t5+2tlimt→−2t4+limt→−24t2+limt→−24▫3. Apply the constant rule to the term limt→−24◦Recall the definition of the constant ruleLimitC,t=C◦This meanslimt→−24=4We can now rewrite the limit as:limt→−2t2+5tt10−6t5+2tlimt→−2t4+limt→−24t2+4▫4. Apply the constant multiple rule to the term limt→−24t2◦Recall the definition of the constant multiple rulelimt→−2Cft=Climt→−2ft◦This means:limt→−24t2=4⋅limt→−2t2We can rewrite the limit as:limt→−2t2+5tt10−6t5+2tlimt→−2t4+4limt→−2t2+4▫5. Apply the power rule◦Recall the definition of the power rulelimt→atn=limt→atnThis gives:limt→−2t2+5tt10−6t5+2tlimt→−2t4+4limt→−2t2+4▫6. Apply the identity rule◦Recall the definition of the identity rulelimt→at=aThis gives:limt→−2t2+5tt10−6t5+2tlimt→−2t4+20▫7. Apply the power rule◦Recall the definition of the power rulelimt→atn=limt→atnThis gives:limt→−2t2+5tt10−6t5+2tlimt→−2t4+20▫8. Apply the identity rule◦Recall the definition of the identity rulelimt→at=aThis gives:limt→−2t2+5tt10−6t5+2t36▫9. Apply the product rule◦Recall the definition of the product rulelimt→aftgt=limt→aftlimt→agtft=t2+5tgt=t10−6t5+2tThis gives:limt→−2t2+5tlimt→−2t10−6t5+2t36▫10. Apply the sum rule◦Recall the definition of the sum rulelimt→aft+gt=limt→aft+limt→agtft=t2gt=5tThis gives:limt→−2t2+limt→−25tlimt→−2t10−6t5+2t36▫11. Apply the constant multiple rule to the term limt→−25t◦Recall the definition of the constant multiple rulelimt→−2Cft=Climt→−2ft◦This means:limt→−25t=5⋅limt→−2tWe can rewrite the limit as:limt→−2t2+5limt→−2tlimt→−2t10−6t5+2t36▫12. Apply the identity rule◦Recall the definition of the identity rulelimt→at=aThis gives:limt→−2t2−10limt→−2t10−6t5+2t36▫13. Apply the power rule◦Recall the definition of the power rulelimt→atn=limt→atnThis gives:limt→−2t2−10limt→−2t10−6t5+2t36▫14. Apply the identity rule◦Recall the definition of the identity rulelimt→at=aThis gives:−limt→−2t10−6t5+2t6▫15. Apply the sum rule◦Recall the definition of the sum rulelimt→af1t+f2t+f3t=limt→af1t+limt→af2t+limt→af3tf1t=t10f2t=−6t5f3t=2tThis gives:−limt→−2t106−limt→−2−6t56−limt→−22t6▫16. Apply the constant multiple rule to the term limt→−2−6t5◦Recall the definition of the constant multiple rulelimt→−2Cft=Climt→−2ft◦This means:limt→−2−6t5=−6⋅limt→−2t5We can rewrite the limit as:−limt→−2t106+limt→−2t5−limt→−22t6▫17. Apply the power rule◦Recall the definition of the power rulelimt→atn=limt→atnThis gives:−limt→−2t106+limt→−2t5−limt→−22t6▫18. Apply the identity rule◦Recall the definition of the identity rulelimt→at=aThis gives:−limt→−2t106−32−limt→−22t6▫19. Apply the constant multiple rule to the term limt→−22t◦Recall the definition of the constant multiple rulelimt→−2Cft=Climt→−2ft◦This means:limt→−22t=2⋅limt→−2tWe can rewrite the limit as:−limt→−2t106−32−limt→−2t3▫20. Apply the identity rule◦Recall the definition of the identity rulelimt→at=aThis gives:−limt→−2t106−943▫21. Apply the power rule◦Recall the definition of the power rulelimt→atn=limt→atnThis gives:−limt→−2t106−943▫22. Apply the identity rule◦Recall the definition of the identity rulelimt→at=aThis gives:−202
Table 1.3.3(b) Maple's stepwise solution via the All Solution Steps option in the Context Panel
Finally, to evaluate the given limit interactively with the Limit Methods tutor, press the following button.
A tutor can be launched from the Tools≻Tutors menu, or from the Context Panel after the appropriate package has been loaded.
To specify a problem in the Limit Methods tutor, note that the top line of this maplet contains fields for the function, variable, limit point, and whether the limit is two-sided (blank) or one-sided (left or right). Press the Start button, then apply limit laws by clicking the corresponding button in the tutor. The menu bars provide a summary of each known rule (Rule Definition), help, and another way to apply rules (Apply the Rule). Note that the selected rule is generally applied to the first possible occurrence; it may be necessary to apply a rule multiple times in succession.
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