Baixe o app para aproveitar ainda mais
Prévia do material em texto
SOLUTIONS TO LIMITS OF FUNCTIONS AS X APPROACHES PLUS OR MINUS INFINITY SOLUTION 13 : = (This is true because the expression approaches and the expression x + 3 approaches as x approaches . The next step follows from the following simple fact. If A is a positive quantity, then = A . ) = = = (You will learn later that the previous step is valid because of the continuity of the square root function.) = (Inside the square root sign lies an indeterminate form. Circumvent it by dividing each term by , the highest power of x inside the square root sign.) = = = (Each of the three expressions , , and approaches 0 as x approaches .) = = = . Click HERE to return to the list of problems. SOLUTION 14 : = (This is true because the expression approaches and the expression x + 3 approaches as x approaches . The next step follows from the following simple fact. If A is a negative quantity, then = A so that = ( A ) = A . Please make sure that you think about and understand this before proceeding. ) = = = (You will learn later that the previous step is valid because of the continuity of the square root function.) = (Inside the square root sign lies an indeterminate form. Circumvent it by dividing each term by , the highest power of x inside the square root sign.) = = = (Each of the three expressions , , and approaches 0 as x approaches .) = = = . Click HERE to return to the list of problems. SOLUTION 15 : = (You will learn later that the previous step is valid because of the continuity of the logarithm function. Note also that the expression leads to the indeterminate form . Circumvent it by dividing each term by , the highest power of x .) = = = (The term approaches 0 as x approaches .) = = = 0 . Click HERE to return to the list of problems. SOLUTION 16 : = (You will learn later that the previous step is valid because of the continuity of the cosine function.) = = (The expression leads to the indeterminate form . Circumvent it by dividing each term by , the highest power of x in the expression.) = = = (Each of the terms and approaches 0 as x approaches .) = = = . Click HERE to return to the list of problems. SOLUTION 17 : (As x approaches each of the expressions and approaches 0. The following steps explain why.) = = = = = 0 . Click HERE to return to the list of problems. SOLUTION 18 : = (Circumvent this indeterminate form by dividing each term in the expression by . Division by also works . You might want to try it both ways to convince yourself of this. Also, BEWARE of making one of the following common MISTAKES : = or \ = .) = = = (Since approaches 0 and approaches as x approaches , we get the following resultant limit.) = = . (Thus, the limit does not exist.) Click HERE to return to the list of problems. SOLUTION 19 : = `` '' truein truein (BEWARE of making the following common MISTAKE : = . Realize also that the form `` '' is an indeterminate one ! It is not equal to 1 ! Circumvent it in the following algebraic ways.) = = (Factor out the term . If you have time, try factoring out the term to convince yourself that it DOESN'T seem to help !) = = = = = (The expressions and approach 0 as x approaches .) = = . = 9 . Click HERE to return to the list of problems.
Compartilhar