Question

discuss the continuity of f(x)= { e^x-1/log(1+2x), x not equal to zero, 7 , when x=0} at x=0

Answer 1

i used standard limits.......it is very simple

Answer 2

The function isn't continuous that is discontinuous

• For checking the continuity of the function at x=0 we have to find the limit of the function when x tends to 0 and whether the limit exists or not and the limiting value is equal to the functional value when x = 0
• Now $\lim_{x \to \ 0} \frac{e^{x}-1 }{log(1+2x)}$ is the limit which we have to calculate
• putting x = 0 we found that this is of $\frac{0}{0}$ form which is indeterminate form
• Hence we can use L'Hospital rule here to calculate the limit

• Upon differentiating w.r.t x we get $\lim_{x \to \ 0} \frac{e^{x} }{\frac{2}{1+2x} }$

• On simplification we get $\lim_{x \to \ 0}\frac{e^{x}(1+2x) }{2}$

• So the limiting value is 1/2

• But the functional value is given as 7. Hence they are not same. So the function is not continuous at x=0